#set document(title: "9.3 The Dot Product", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 9.3#h(0.6em)The Dot Product === Components We have seen that it can be useful to resolve a vector into horizontal and vertical components. We can also break a vector into components that point in other directions. Imagine the following experiment: Delbert holds a ball at shoulder height and then drops it, so that it falls to the ground. Francine holds a ball at shoulder height on an inclined ramp, then releases it so that it rolls downhill. Which ball will reach the ground first? Although gravity causes both balls to speed up, the free-falling ball will reach the ground first. The force of gravity pulls straight down, the same direction as the motion of the free-falling ball, but the rolling ball must move at an angle to the pull of gravity, along the surface of the ramp. Only part of the gravitational force accelerates the rolling ball, and the rest of the force is counteracted by the surface of the ramp. What fraction of the gravitational force causes the ball to roll? #figure(figph[ramp], alt: "ramp", caption: none) In figure (a), the gravitational force #math.equation(block: false, alt: "F")[$bold(F)$] is resolved into the sum of two vectors, #math.equation(block: false, alt: "F equals u plus v")[$bold(F) = bold(u) + bold(v)$], where #math.equation(block: false, alt: "v")[$bold(v)$] points down the ramp, and #math.equation(block: false, alt: "u")[$bold(u)$] is perpendicular to the ramp. The magnitude of #math.equation(block: false, alt: "v")[$bold(v)$] is called the component of #math.equation(block: false, alt: "F")[$bold(F)$] in the direction of motion, and is denoted by #math.equation(block: false, alt: "comp sub v F")[$"comp"_(bold(v)) bold(F)$]. This is the portion of the gravitational force that moves the ball. From figure (b), we see that #math.equation(block: false, alt: "comp sub v F equals ∥ F ∥ cos ⁡ open parenthesis θ close parenthesis")[$"comp"_(bold(v)) bold(F) = ∥ bold(F) ∥ cos ( θ )$], where #math.equation(block: false, alt: "θ")[$θ$] is the angle between #math.equation(block: false, alt: "F")[$bold(F)$] and #math.equation(block: false, alt: "v")[$bold(v)$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Component of a Vector] The component of a vector #math.equation(block: false, alt: "F")[$bold(F)$] in the direction of vector #math.equation(block: false, alt: "v")[$bold(v)$] is #math.equation(block: true, alt: "comp sub v F equals ∥ F ∥ cos ⁡ open parenthesis θ close parenthesis")[$"comp"_(bold(v)) bold(F) = ∥ bold(F) ∥ bold(cos) ( bold(italic(θ)) )$] where #math.equation(block: false, alt: "θ")[$θ$] is the angle between #math.equation(block: false, alt: "F")[$bold(F)$] and #math.equation(block: false, alt: "v")[$bold(v)$]. ] With a little geometry, you can verify that in this example the angle #math.equation(block: false, alt: "θ")[$θ$] is the complement of the angle of inclination of the ramp, #math.equation(block: false, alt: "α")[$α$]. (Think of similar triangles.) Now suppose that we increase the angle of inclination. As #math.equation(block: false, alt: "α")[$α$] increases, #math.equation(block: false, alt: "θ")[$θ$] decreases, #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$] increases, and hence #math.equation(block: false, alt: "comp sub v F")[$"comp"_(bold(v)) bold(F)$] increases. This result agrees with our experience: as the ramp gets steeper, the ball rolls faster. #examplebox("Example 1")[][ Delbert uses a sheet of plywood as a ramp for his wheelbarrow. The ramp is inclined at an angle of #math.equation(block: false, alt: "15 degrees")[$15^(∘)$] to the horizontal. The plywood can support a maximum weight of 200 pounds. What weight will the ramp support? #solutionbox[ The component of the weight perpendicular to the ramp cannot be more than 200 pounds. If the weight of the wheelbarrow is #math.equation(block: false, alt: "w")[$bold(w)$], and #math.equation(block: false, alt: "v")[$bold(v)$] is a vector perpendicular to the ramp, as shown below, then the angle between #math.equation(block: false, alt: "v")[$bold(v)$] and #math.equation(block: false, alt: "w")[$bold(w)$] is #math.equation(block: false, alt: "15 degrees")[$15^(∘)$], by similar triangles. Thus, the component of #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of #math.equation(block: false, alt: "w")[$bold(w)$] is #math.equation(block: true, alt: "comp sub v w equals ∥ w ∥ cos ⁡ open parenthesis 15 degrees close parenthesis, equals 200; ∥ w ∥ open parenthesis 0.9659 close parenthesis, equals 200; ∥ w ∥, equals 207.06")[$"comp"_(bold(v)) bold(w) = ∥ bold(w) ∥ cos ( 15^(∘) ) & = 200 \ ∥ bold(w) ∥ ( 0.9659 ) & = 200 \ ∥ bold(w) ∥ & = 207.06$] #figure(figph[ramp], alt: "ramp", caption: none) The ramp can support no more than about 207 pounds. ] ] Delbert pushes a lawn mower whose handle makes a #math.equation(block: false, alt: "40 degrees")[$40^(∘)$] angle with the horizontal. If he applies a force of 30 pounds in the direction of the handle, what is the component of the force in the horizontal direction? #figure(figph[lawn mower], alt: "lawn mower", caption: none) 23 pounds === Coordinate Form for Components In the examples above, we computed the component of a force #math.equation(block: false, alt: "F")[$bold(F)$] in the direction of a vector #math.equation(block: false, alt: "v")[$bold(v)$] by knowing the angle between #math.equation(block: false, alt: "F")[$bold(F)$] and #math.equation(block: false, alt: "v")[$bold(v)$]. If the vectors are given in coordinate form (that is, #math.equation(block: false, alt: "v equals a i plus b j")[$bold(v) = a bold(i) + b bold(j)$]), we may not know the angle between them. Can we compute the component of a vector #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of #math.equation(block: false, alt: "v")[$bold(v)$], in terms of the coordinates of #math.equation(block: false, alt: "v")[$bold(v)$] and#math.equation(block: false, alt: "w")[$bold(w)$]? Suppose #math.equation(block: false, alt: "v equals a i plus b j")[$bold(v) = a bold(i) + b bold(j)$] and #math.equation(block: false, alt: "w equals c i plus d j")[$bold(w) = c bold(i) + d bold(j)$], as shown below. We will need to compute the cosine of #math.equation(block: false, alt: "θ")[$θ$] in terms of #math.equation(block: false, alt: "a , b , c ,")[$a , " " b , " " c , " "$] and #math.equation(block: false, alt: "d")[$d$]. First verify that #math.equation(block: true, alt: "cos ⁡ open parenthesis α close parenthesis, equals the fraction a over ∥ v ∥ and sin ⁡ open parenthesis α close parenthesis equals the fraction b over ∥ v ∥; cos ⁡ open parenthesis β close parenthesis, equals the fraction c over ∥ w ∥ and sin ⁡ open parenthesis β close parenthesis equals the fraction d over ∥ w ∥")[$cos ( α ) & = display(frac(a, ∥ bold(v) ∥)) " " " " " " "and" " " " " " " sin ( α ) = display(frac(b, ∥ bold(v) ∥)) \ cos ( β ) & = display(frac(c, ∥ bold(w) ∥)) " " " " " " "and" " " " " " " sin ( β ) = display(frac(d, ∥ bold(w) ∥))$] #figure(figph[vectors and components], alt: "vectors and components", caption: none) Because #math.equation(block: false, alt: "θ equals β equals α")[$θ = β = α$], we use the subtraction formula for cosine. #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis, equals cos ⁡ open parenthesis α close parenthesis cos ⁡ open parenthesis β close parenthesis plus sin ⁡ open parenthesis α close parenthesis sin ⁡ open parenthesis β close parenthesis; equals the fraction a over ∥ v ∥ the fraction c over ∥ w ∥ plus the fraction b over ∥ v ∥ the fraction d over ∥ w ∥; equals the fraction 1 over ∥ v ∥ ∥ w ∥ open parenthesis a c plus b d close parenthesis")[$cos ( θ ) & = cos ( α ) cos ( β ) + sin ( α ) sin ( β ) \ & = display(frac(a, ∥ bold(v) ∥)) display(frac(c, ∥ bold(w) ∥)) + display(frac(b, ∥ bold(v) ∥)) display(frac(d, ∥ bold(w) ∥)) \ & = display(frac(1, ∥ bold(v) ∥ ∥ bold(w) ∥)) ( a c + b d )$] And finally, #math.equation(block: true, alt: "comp sub v w equals ∥ w ∥ cos ⁡ open parenthesis θ close parenthesis equals the fraction a c plus b d over ∥ v ∥")[$"comp"_(bold(v)) bold(w) = ∥ bold(w) ∥ cos ( θ ) = display(frac(a c + b d, ∥ bold(v) ∥))$] Now we have a fromula for the component of a vector #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of a vector #math.equation(block: false, alt: "v")[$bold(v)$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Component of a Vector in Coordinate Form] If #math.equation(block: false, alt: "v equals a i plus b j")[$" " bold(v) = a bold(i) + b bold(j) " "$] and #math.equation(block: false, alt: "w equals c i plus d j")[$" " bold(w) = c bold(i) + d bold(j)$], then the component of #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of #math.equation(block: false, alt: "v")[$bold(v)$] is given by #math.equation(block: true, alt: "comp sub v w equals the fraction a c plus b d over ∥ v ∥")[$"comp"_(bold(v)) bold(w) = display(frac(bold(italic(a)) bold(italic(c)) + bold(italic(b)) bold(italic(d)), ∥ bold(v) ∥))$] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Recall that the "component" of a vector is a scalar. If a vector #math.equation(block: false, alt: "w")[$bold(w)$] is resolved into two component vectors, #math.equation(block: false, alt: "w equals u plus v")[$bold(w) = bold(u) + bold(v)$], its components are the #emph[lengths] of the vectors #math.equation(block: false, alt: "u")[$bold(u)$] and #math.equation(block: false, alt: "v")[$bold(v)$]. The vectors #math.equation(block: false, alt: "u")[$bold(u)$] and #math.equation(block: false, alt: "v")[$bold(v)$] themselves are called the #strong[vector components] of #math.equation(block: false, alt: "w")[$bold(w)$]. ] #examplebox("Example 2")[][ + Compute the component of #math.equation(block: false, alt: "w equals 6 i plus 2 j")[$bold(w) = 6 bold(i) + 2 bold(j)$] in the direction of #math.equation(block: false, alt: "v equals 4 i plus 3 j")[$bold(v) = 4 bold(i) + 3 bold(j)$]. + Find the vector component of #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of #math.equation(block: false, alt: "v")[$bold(v)$]. + Find the vector component of #math.equation(block: false, alt: "w")[$bold(w)$] perpendicular to #math.equation(block: false, alt: "v")[$bold(v)$]. #solutionbox[ + We use the formula derived above. #math.equation(block: true, alt: "comp sub v w equals the fraction a c plus b d over ∥ v ∥ equals the fraction 4 open parenthesis 6 close parenthesis plus 3 open parenthesis 2 close parenthesis over the square root of 4 squared plus 3 squared equals the fraction 30 over 5 equals 6")[$"comp"_(bold(v)) bold(w) = display(frac(a c + b d, ∥ bold(v) ∥)) = display(frac(4 ( 6 ) + 3 ( 2 ), sqrt(4^(2) + 3^(2)))) = display(frac(30, 5)) = 6$] Remember that the component of #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of #math.equation(block: false, alt: "v")[$bold(v)$] is a scalar; it is the length of the projection of #math.equation(block: false, alt: "w")[$bold(w)$] onto #math.equation(block: false, alt: "v")[$bold(v)$]. + We would like to find a vector #math.equation(block: false, alt: "q")[$bold(q)$] of length 6 in the direction of #math.equation(block: false, alt: "v")[$bold(v)$], as shown at right. Because #math.equation(block: false, alt: "∥ v ∥ equals 5")[$∥ bold(v) ∥ = 5$], we scale the components of #math.equation(block: false, alt: "v")[$bold(v)$] by #math.equation(block: false, alt: "the fraction 6 over 5")[$display(frac(6, 5))$] to get #math.equation(block: true, alt: "q equals the fraction 6 over 5 open parenthesis 4 i plus 3 j close parenthesis equals the fraction 24 over 5 i plus the fraction 18 over 5 j")[$bold(q) = display(frac(6, 5)) ( 4 bold(i) + 3 bold(j) ) = display(frac(24, 5)) bold(i) + display(frac(18, 5)) bold(j)$]#figure(figph[vectors], alt: "vectors", caption: none) + Because #math.equation(block: false, alt: "w equals p plus q")[$bold(w) = bold(p) + bold(q)$], where #math.equation(block: false, alt: "p")[$bold(p)$] is perpendicular to #math.equation(block: false, alt: "q")[$bold(q)$], we have #math.equation(block: true, alt: "p equals w minus q equals open parenthesis 6 i plus 2 j close parenthesis minus open parenthesis the fraction 24 over 5 i plus the fraction 18 over 5 j close parenthesis equals the fraction 6 over 5 i minus the fraction 8 over 5 j")[$bold(p) = bold(w) − bold(q) = ( 6 bold(i) + 2 bold(j) ) − ( display(frac(24, 5)) bold(i) + display(frac(18, 5)) bold(j) ) = display(frac(6, 5)) bold(i) − display(frac(8, 5)) bold(j)$] ] ] Write the vector #math.equation(block: false, alt: "w equals 6 i plus 2 j")[$bold(w) = 6 bold(i) + 2 bold(j)$] as the sum of two components, one parallel to #math.equation(block: false, alt: "v equals i plus j")[$bold(v) = bold(i) + bold(j)$] and the other perpendicular to #math.equation(block: false, alt: "v")[$bold(v)$]. The example's decomposition, live. The dashed line carries the direction of v = i + j; the blue vector is w = ci + dj with its components on sliders; the red vector is the part of w parallel to v, and the gray dashed connector from its tip to w's tip is the perpendicular part — at every setting the two add back to w, and the connector meets the dashed line at a right angle. At the starting values w = 6i + 2j, this is exactly the worked example: w = (4i + 4j) + (2i − 2j). The red part's length is the component of w in the direction of v — and its sign matters: slide w below the dashed line's perpendicular (make c + d negative, say w = i − 4j) and the red vector points backward along v, a negative component. At c + d = 0 — try w = 3i − 3j — w is perpendicular to v, the parallel part vanishes entirely, and w is all crosswise: the orthogonal case, where the dot product is zero.{"functions":\[{"color":"\#6a6a6a","expression":"t","expression2":"t","id":"v-direction","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"parametric","tMax":6,"tMin":-3.2,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"t\*c","expression2":"t\*d","id":"vector-w","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"t\*(c + d)/2","expression2":"t\*(c + d)/2","id":"parallel-part","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#6a6a6a","expression":"(c + d)/2 + t\*(c - (c + d)/2)","expression2":"(c + d)/2 + t\*(d - (c + d)/2)","id":"perp-part","inequality":"lt","lineDash":\[4,4\],"lineWidth":2,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"c + 0.13\*cos(t)","expression2":"d + 0.13\*sin(t)","id":"w-dot","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-c","label":"Component c of w (east-west)","max":7,"min":1,"name":"c","step":1,"value":6},{"id":"p-d","label":"Component d of w (north-south)","max":4,"min":-4,"name":"d","step":1,"value":2}\],"title":"w = ci + dj split along v = i + j and across it","version":1,"viewport":{"centerX":2.5,"centerY":1,"scale":40}} #math.equation(block: true, alt: "w equals open parenthesis 4 i plus 4 j close parenthesis plus open parenthesis 2 i minus 2 j close parenthesis")[$bold(w) = ( 4 bold(i) + 4 bold(j) ) + ( 2 bold(i) − 2 bold(j) )$] === The Dot Product The expression #math.equation(block: false, alt: "a c plus b d")[$a c + b d$], which we encountered above as part of the formula for #math.equation(block: false, alt: "comp sub v w")[$"comp"_(bold(v)) bold(w)$], is quite useful and is given a name; it is called the #strong[dot product] of the vectors #math.equation(block: false, alt: "v equals a i plus b j")[$bold(v) = a bold(i) + b bold(j)$] and #math.equation(block: false, alt: "w equals c i plus d j")[$bold(w) = c bold(i) + d bold(j)$]. It is easy to remember the formula for the dot product if we think of adding the product of the #math.equation(block: false, alt: "i")[$bold(i)$]-components and the product of the #math.equation(block: false, alt: "j")[$bold(j)$]-components of the two vectors. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Dot Product] The #strong[dot product] of two vectors #math.equation(block: false, alt: "v equals v sub 1 i plus v sub 2 j")[$" " bold(v) = v_(1) bold(i) + v_(2) bold(j) " "$] and #math.equation(block: false, alt: "w equals w sub 1 i plus w sub 2 j")[$" " bold(w) = w_(1) bold(i) + w_(2) bold(j) " "$] is the scalar #math.equation(block: true, alt: "v times w equals v sub 1 w sub 1 plus v sub 2 w sub 2")[$bold(v) ⋅ bold(w) = bold(italic(v))_(1) bold(italic(w))_(1) + bold(italic(v))_(2) bold(italic(w))_(2)$] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ From the formula above, you can see that the dot product is commutative. That is, #math.equation(block: true, alt: "v times w equals w times v")[$bold(v) ⋅ bold(w) = bold(w) ⋅ bold(v)$] You can check that this is true in the following example. ] #examplebox("Example 3")[][ Compute the dot product of #math.equation(block: false, alt: "v equals 5 i minus 3 j")[$bold(v) = 5 bold(i) − 3 bold(j)$] and #math.equation(block: false, alt: "w equals 4 i plus j")[$bold(w) = 4 bold(i) + bold(j)$]. #solutionbox[ We apply the formula above to find #math.equation(block: true, alt: "v times w equals 5 open parenthesis 4 close parenthesis plus open parenthesis minus 3 close parenthesis open parenthesis 1 close parenthesis equals 17")[$bold(v) ⋅ bold(w) = 5 ( 4 ) + ( − 3 ) ( 1 ) = 17$] ] ] In the examples above, you can see that the dot product of two vectors is a scalar. For this reason, the dot product is also called the #strong[scalar product]. Compute the dot product of #math.equation(block: false, alt: "v equals 6 i plus 2 j")[$bold(v) = 6 bold(i) + 2 bold(j)$] and #math.equation(block: false, alt: "w equals minus 2 i plus 3 j")[$bold(w) = − 2 bold(i) + 3 bold(j)$]. #math.equation(block: true, alt: "minus 6")[$− 6$] We can now write the formula for #math.equation(block: false, alt: "comp sub v w")[$"comp"_(bold(v)) bold(w)$] using the dot product. #math.equation(block: true, alt: "comp sub v w equals ∥ w ∥ cos ⁡ open parenthesis θ close parenthesis equals the fraction a c plus b d over ∥ v ∥ equals the fraction v times w over ∥ v ∥")[$"comp"_(bold(v)) bold(w) = ∥ bold(w) ∥ cos ( θ ) = display(frac(a c + b d, ∥ bold(v) ∥)) = display(frac(bold(v) ⋅ bold(w), ∥ bold(v) ∥))$] We have derived an alternate formula for a component of a vector. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Component of a Vector] The component of #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of #math.equation(block: false, alt: "v")[$bold(v)$] is the scalar #math.equation(block: true, alt: "comp sub v w equals the fraction v times w over ∥ v ∥")[$"comp"_(bold(v)) bold(w) = display(frac(bold(v) ⋅ bold(w), ∥ bold(v) ∥))$] ] #examplebox("Example 4")[][ Compute the component of #math.equation(block: false, alt: "v equals 5 i minus 3 j")[$bold(v) = 5 bold(i) − 3 bold(j)$] in the direction of #math.equation(block: false, alt: "w equals 4 i plus j")[$bold(w) = 4 bold(i) + bold(j)$]. #solutionbox[ Using the formula above, we have #math.equation(block: false, alt: "comp sub w v equals the fraction v times w over ∥ w ∥")[$"comp"_(bold(w)) bold(v) = display(frac(bold(v) ⋅ bold(w), ∥ bold(w) ∥))$]. We first compute #math.equation(block: false, alt: "v times w")[$bold(v) ⋅ bold(w)$] and #math.equation(block: false, alt: "∥ w ∥")[$∥ bold(w) ∥$]. #math.equation(block: true, alt: "v times w equals 5 open parenthesis 4 close parenthesis plus open parenthesis minus 3 close parenthesis open parenthesis 1 close parenthesis equals 17 and ∥ w ∥ equals the square root of 4 squared plus 1 squared equals the square root of 17")[$bold(v) ⋅ bold(w) = 5 ( 4 ) + ( − 3 ) ( 1 ) = 17 " " " " "and" " " " " ∥ bold(w) ∥ = sqrt(4^(2) + 1^(2)) = sqrt(17)$] Thus, #math.equation(block: true, alt: "comp sub w v equals the fraction v times w over ∥ w ∥ equals the fraction 17 over the square root of 17 equals the square root of 17")[$"comp"_(bold(w)) bold(v) = display(frac(bold(v) ⋅ bold(w), ∥ bold(w) ∥)) = display(frac(17, sqrt(17))) = sqrt(17)$] The length of the projection of #math.equation(block: false, alt: "v")[$bold(v)$] in the direction of #math.equation(block: false, alt: "w")[$bold(w)$] is #math.equation(block: false, alt: "the square root of 17")[$sqrt(17)$] units, as shown at right. #figure(figph[vectors on grid], alt: "vectors on grid", caption: none) ] ] Compute the component of #math.equation(block: false, alt: "u equals 2 i plus 3 j")[$bold(u) = 2 bold(i) + 3 bold(j)$] in the direction of #math.equation(block: false, alt: "v equals 6 i plus 5 j")[$bold(v) = 6 bold(i) + 5 bold(j)$]. #math.equation(block: true, alt: "the fraction 27 over the square root of 61")[$display(frac(27, sqrt(61)))$] === Geometric Meaning of the Dot Product An even more important relationship, which gives geometric meaning to the dot product, follows from the formula for a component. We now have two ways to compute the component of #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of #math.equation(block: false, alt: "v")[$bold(v)$]: #math.equation(block: true, alt: "comp sub v w equals ∥ w ∥ cos ⁡ θ and comp sub v w equals the fraction v times w over ∥ v ∥")[$"comp"_(bold(v)) bold(w) = ∥ bold(w) ∥ cos θ " " " " " " " " " " " " "and" " " " " " " " " " " " " "comp"_(bold(v)) bold(w) = display(frac(bold(v) ⋅ bold(w), ∥ bold(v) ∥))$] Equating these two expressions, we find #math.equation(block: true, alt: "∥ w ∥ cos ⁡ open parenthesis θ close parenthesis equals the fraction v times w over ∥ v ∥")[$∥ bold(w) ∥ cos ( θ ) = display(frac(bold(v) ⋅ bold(w), ∥ bold(v) ∥))$] or #math.equation(block: false, alt: "∥ v ∥ ∥ w ∥ cos ⁡ θ equals v times w")[$∥ bold(v) ∥ ∥ bold(w) ∥ cos θ = bold(v) ⋅ bold(w)$]. This is a geometric formula for the dot product. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Geometric Form for the Dot Product] The #strong[dot product] of two vectors #math.equation(block: false, alt: "v")[$bold(v)$] and #math.equation(block: false, alt: "w")[$bold(w)$] is the scalar #math.equation(block: true, alt: "v times w equals ∥ v ∥ ∥ w ∥ cos ⁡ open parenthesis θ close parenthesis")[$bold(v) ⋅ bold(w) = ∥ bold(v) ∥ ∥ bold(w) ∥ bold(cos) ( bold(italic(θ)) )$] where #math.equation(block: false, alt: "θ")[$θ$] is the angle between the vectors. ] The dot product is a way of multiplying two vectors that depends on the angle between them. - If #math.equation(block: false, alt: "θ equals 0 degrees")[$θ = 0^(∘)$], so that #math.equation(block: false, alt: "v")[$bold(v)$] and #math.equation(block: false, alt: "w")[$bold(w)$] point in the same direction, then #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 1")[$cos ( θ ) = 1$] and #math.equation(block: false, alt: "v times w")[$bold(v) ⋅ bold(w)$] is just the product of their lengths, #math.equation(block: false, alt: "∥ v ∥ ∥ w ∥")[$∥ bold(v) ∥ ∥ bold(w) ∥$]. - If #math.equation(block: false, alt: "v")[$bold(v)$] and #math.equation(block: false, alt: "w")[$bold(w)$] are perpendicular, then #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 0")[$cos ( θ ) = 0$], so #math.equation(block: false, alt: "v times w equals 0")[$bold(v) ⋅ bold(w) = 0$]. (Two vectors #math.equation(block: false, alt: "v")[$bold(v)$] and #math.equation(block: false, alt: "w")[$bold(w)$] are said to be #strong[orthogonal] if their dot product is zero.) - If #math.equation(block: false, alt: "θ")[$θ$] is between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "90 degrees")[$90^(∘)$], the dot product multiplies the length of #math.equation(block: false, alt: "v")[$bold(v)$] times the component of #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of #math.equation(block: false, alt: "v")[$bold(v)$]. #figure(figph[vectors], alt: "vectors", caption: none) #examplebox("Example 5")[][ Show that the vectors #math.equation(block: false, alt: "v equals 2 i plus 6 j")[$" " bold(v) = 2 bold(i) + 6 bold(j) " "$] and #math.equation(block: false, alt: "w equals minus 9 i plus 3 j")[$" " bold(w) = − 9 bold(i) + 3 bold(j) " "$] are orthogonal. #solutionbox[ We compute the dot product of the vectors. #math.equation(block: true, alt: "v times w equals 2 open parenthesis minus 9 close parenthesis plus 6 open parenthesis 3 close parenthesis equals 0")[$bold(v) ⋅ bold(w) = 2 ( − 9 ) + 6 ( 3 ) = 0$] Thus, #math.equation(block: false, alt: "∥ v ∥ ∥ w ∥ cos ⁡ open parenthesis θ close parenthesis equals 0")[$∥ bold(v) ∥ ∥ bold(w) ∥ cos ( θ ) = 0$], so #math.equation(block: false, alt: "v")[$bold(v)$] and #math.equation(block: false, alt: "w")[$bold(w)$] are orthogonal. Because neither #math.equation(block: false, alt: "∥ v ∥ equals 0")[$∥ bold(v) ∥ = 0$] or #math.equation(block: false, alt: "∥ w ∥ equals 0")[$∥ bold(w) ∥ = 0$], it must be the case that #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 0")[$cos ( θ ) = 0$], so #math.equation(block: false, alt: "θ equals 90 degrees")[$θ = 90^(∘)$] or #math.equation(block: false, alt: "270 degrees")[$270^(∘)$]. ] ] + Show that the vectors #math.equation(block: false, alt: "v equals a i plus b j")[$bold(v) = a bold(i) + b bold(j)$] and #math.equation(block: false, alt: "w equals minus b i plus a j")[$bold(w) = − b bold(i) + a bold(j)$] are orthogonal. + Find a vector #math.equation(block: false, alt: "w")[$bold(w)$] perpendicular to #math.equation(block: false, alt: "v equals minus 3 i minus 5 j")[$bold(v) = − 3 bold(i) − 5 bold(j)$]. + #math.equation(block: false, alt: "u times v equals minus a b plus a b equals 0")[$bold(u) ⋅ bold(v) = − a b + a b = 0$] + #math.equation(block: false, alt: "5 i minus 3 j")[$5 bold(i) − 3 bold(j)$] Using the dot product, we can find the angle between two vectors. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Angle Between Two Vectors] The angle #math.equation(block: false, alt: "θ")[$θ$] between two vectors #math.equation(block: false, alt: "v")[$bold(v)$] and #math.equation(block: false, alt: "w")[$bold(w)$] is given by #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction v times w over ∥ v ∥ ∥ w ∥")[$bold(cos) ( bold(italic(θ)) ) = display(frac(bold(v) ⋅ bold(w), ∥ bold(v) ∥ ∥ bold(w) ∥))$] ] #examplebox("Example 6")[][ Find the angle between the vectors #math.equation(block: false, alt: "u equals minus 3 i plus 2 j")[$bold(u) = − 3 bold(i) + 2 bold(j)$] and #math.equation(block: false, alt: "v equals 5 i plus 3 j")[$bold(v) = 5 bold(i) + 3 bold(j)$]. #solutionbox[ We first compute #math.equation(block: false, alt: "u times v")[$bold(u) ⋅ bold(v)$] and the magnitude of each vector. #math.equation(block: true, alt: "u times v, equals minus 3 open parenthesis 5 close parenthesis plus 2 open parenthesis 3 close parenthesis equals minus 9; ∥ u ∥, equals the square root of open parenthesis minus 3 close parenthesis squared plus 2 squared equals the square root of 13; ∥ v ∥, equals the square root of 5 squared plus 3 squared equals the square root of 34")[$bold(u) ⋅ bold(v) & = − 3 ( 5 ) + 2 ( 3 ) = − 9 \ ∥ bold(u) ∥ & = sqrt(( − 3 )^(2) + 2^(2)) = sqrt(13) \ ∥ bold(v) ∥ & = sqrt(5^(2) + 3^(2)) = sqrt(34)$] #figure(figph[vectors], alt: "vectors", caption: none) Then we apply the formula for #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis")[$cos ( θ )$]. #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis, equals the fraction v times w over ∥ v ∥ ∥ w ∥ equals the fraction minus 9 over the square root of 13 the square root of 34 equals minus 0.4281; θ, equals cos to the power minus 1 ⁡ open parenthesis minus 0.4281 close parenthesis equals 115.35 degrees")[$cos ( θ ) & = display(frac(bold(v) ⋅ bold(w), ∥ bold(v) ∥ ∥ bold(w) ∥)) = display(frac(− 9, sqrt(13) sqrt(34))) = − 0.4281 \ θ & = cos^(− 1) ( − 0.4281 ) = 115.35^(∘)$] The angle between the vectors is #math.equation(block: false, alt: "115.35 degrees")[$115.35^(∘)$]. ] ] Find the angle between the vectors #math.equation(block: false, alt: "u equals 4 i minus 6 j")[$bold(u) = 4 bold(i) − 6 bold(j)$] and #math.equation(block: false, alt: "v equals 2 i plus 8 j")[$bold(v) = 2 bold(i) + 8 bold(j)$]. #math.equation(block: true, alt: "132.27 degrees")[$132.27^(∘)$] Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + + Find the height #math.equation(block: false, alt: "h")[$h$] of the triangle. + How far is the foot of the altitude from the vertex of the #math.equation(block: false, alt: "18 degrees")[$18^(∘)$] angle? #figure(figph[triangle], alt: "triangle", caption: none) + + Find the height #math.equation(block: false, alt: "h")[$h$] of the triangle. + Find the length of the third side of the triangle. #figure(figph[triangle], alt: "triangle", caption: none) + + How far north is the tower from the airport? How far east? + What is the distance from #math.equation(block: false, alt: "A")[$A$] to #math.equation(block: false, alt: "P")[$P$]? #figure(figph[triangle], alt: "triangle", caption: none) + + Find the distance from #math.equation(block: false, alt: "A")[$A$] to #math.equation(block: false, alt: "C")[$C$]. + How far north is point #math.equation(block: false, alt: "C")[$C$] from point #math.equation(block: false, alt: "B")[$B$]? #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Skills Refresher Answers + + #math.equation(block: false, alt: "1.55")[$1.55$] + #math.equation(block: false, alt: "4.76")[$4.76$] + + #math.equation(block: false, alt: "19.57")[$19.57$] + #math.equation(block: false, alt: "81.83")[$81.83$] + + 36.82 mi, 15.63 mi + 16.08 mi + + #math.equation(block: false, alt: "21.87")[$21.87$] + #math.equation(block: false, alt: "4.93")[$4.93$] ] === Section 9.3 Summary ==== Vocabulary - Dot product - Scalar product - Orthogonal ==== Concepts + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Dot Product (Coordinate Formula)] The #strong[dot product] of two vectors #math.equation(block: false, alt: "v equals v sub 1 i plus v sub 2 j")[$bold(v) = v_(1) bold(i) + v_(2) bold(j)$] and #math.equation(block: false, alt: "w equals w sub 1 i plus w sub 2 j")[$bold(w) = w_(1) bold(i) + w_(2) bold(j)$] is the scalar #math.equation(block: true, alt: "v times w equals v sub 1 w sub 1 plus v sub 2 w sub 2")[$bold(v) ⋅ bold(w) = v_(1) w_(1) + v_(2) w_(2)$] ] + The dot product is a way of multiplying two vectors that depends on the angle between them. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Dot Product (Geometric Formula)] The #strong[dot product] of two vectors #math.equation(block: false, alt: "v")[$bold(v)$] and #math.equation(block: false, alt: "w")[$bold(w)$] is the scalar #math.equation(block: true, alt: "v times w equals ∥ v ∥ ∥ w ∥ cos ⁡ open parenthesis θ close parenthesis")[$bold(v) ⋅ bold(w) = ∥ bold(v) ∥ ∥ bold(w) ∥ cos ( θ )$]where #math.equation(block: false, alt: "θ")[$θ$] is the angle between the vectors. ] + The component of a vector #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of vector #math.equation(block: false, alt: "v")[$bold(v)$] is the length of the vector projection of #math.equation(block: false, alt: "w")[$bold(w)$] onto #math.equation(block: false, alt: "w")[$bold(w)$]. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Component of a Vector] The component of #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of #math.equation(block: false, alt: "v")[$bold(v)$] is the scalar #math.equation(block: true, alt: "comp sub v w equals the fraction v times w over ∥ v ∥")[$"comp"_(bold(v)) bold(w) = display(frac(bold(v) ⋅ bold(w), ∥ bold(v) ∥))$] ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Angle Between Two Vectors] The angle #math.equation(block: false, alt: "θ")[$θ$] between two vectors #math.equation(block: false, alt: "v")[$bold(v)$] and #math.equation(block: false, alt: "w")[$bold(w)$] is given by #math.equation(block: true, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction v times w over ∥ v ∥ ∥ w ∥")[$cos ( θ ) = display(frac(bold(v) ⋅ bold(w), ∥ bold(v) ∥ ∥ bold(w) ∥))$] ] + Two vectors #math.equation(block: false, alt: "v")[$bold(v)$] and #math.equation(block: false, alt: "w")[$bold(w)$] are orthogonal if #math.equation(block: false, alt: "v times w equals 0")[$bold(v) ⋅ bold(w) = 0$] ==== Study Questions + If #math.equation(block: false, alt: "u")[$bold(u)$] and #math.equation(block: false, alt: "v")[$bold(v)$] have the same direction, what is #math.equation(block: false, alt: "comp sub u v")[$"comp"_(bold(u)) bold(v)$]? + If #math.equation(block: false, alt: "u")[$bold(u)$] is perpendicular to #math.equation(block: false, alt: "v")[$bold(v)$], what is #math.equation(block: false, alt: "comp sub u v")[$"comp"_(bold(u)) bold(v)$]? + What is the angle between #math.equation(block: false, alt: "u")[$bold(u)$] and #math.equation(block: false, alt: "v")[$bold(v)$] that makes their dot product as large as possible? + What does the dot product of two unit vectors tell you? ==== Skills + Find the component of #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of #math.equation(block: false, alt: "v")[$bold(v)$] \#1–6, 37–40 + Compute the dot product \#11–22, 27–36 + Find the angle between two vectors \#23–26 + Resolve a vector into components in given directions \#7–10, 41–42 === Homework 9-3 For Problems 1–6, find the component of #math.equation(block: false, alt: "w")[$bold(w)$] in the direction of #math.equation(block: false, alt: "v")[$bold(v)$]. #math.equation(block: true, alt: "w equals 5 i plus 9 j ,")[$bold(w) = 5 bold(i) + 9 bold(j) ,$] #math.equation(block: true, alt: "v equals 3 i plus 2 j")[$" " " " bold(v) = 3 bold(i) + 2 bold(j)$] #math.equation(block: true, alt: "the fraction 33 over the square root of 13")[$display(frac(33, sqrt(13)))$] #math.equation(block: true, alt: "w equals 7 i plus 4 j ,")[$bold(w) = 7 bold(i) + 4 bold(j) ,$] #math.equation(block: true, alt: "v equals 2 i plus 3 j")[$" " " " bold(v) = 2 bold(i) + 3 bold(j)$] #math.equation(block: true, alt: "w equals minus 6 i plus 5 j ,")[$bold(w) = − 6 bold(i) + 5 bold(j) ,$] #math.equation(block: true, alt: "v equals i plus j")[$" " " " bold(v) = bold(i) + bold(j)$] #math.equation(block: true, alt: "the fraction minus 1 over the square root of 2")[$display(frac(− 1, sqrt(2)))$] #math.equation(block: true, alt: "w equals 10 i minus 14 j , v equals i plus j")[$bold(w) = 10 bold(i) − 14 bold(j) , " " " " bold(v) = bold(i) + bold(j)$] #math.equation(block: true, alt: "w equals 4 i minus 3 j ,")[$bold(w) = 4 bold(i) − 3 bold(j) ,$] #math.equation(block: true, alt: "v equals minus i plus 2 j")[$" " " " bold(v) = − bold(i) + 2 bold(j)$] #math.equation(block: true, alt: "minus 2 the square root of 5")[$− 2 sqrt(5)$] #math.equation(block: true, alt: "w equals minus 2 i minus 3 j , v equals i minus 2 j")[$bold(w) = − 2 bold(i) − 3 bold(j) , " " " " bold(v) = bold(i) − 2 bold(j)$] For Problems 7–10, + Resolve #math.equation(block: false, alt: "w")[$bold(w)$] into two components, one parallel to #math.equation(block: false, alt: "v")[$bold(v)$] and the other orthogonal to #math.equation(block: false, alt: "v")[$bold(v)$]. + Sketch both vectors and the vector components. #math.equation(block: true, alt: "w equals 8 i plus 4 j ,")[$bold(w) = 8 bold(i) + 4 bold(j) ,$] #math.equation(block: true, alt: "v equals 2 i plus 3 j")[$" " " " bold(v) = 2 bold(i) + 3 bold(j)$] #figure(figph[8x8 grid], alt: "8x8 grid", caption: none) + #math.equation(block: false, alt: "w equals open parenthesis the fraction 56 over 13 i plus the fraction 84 over 13 j close parenthesis plus open parenthesis the fraction 48 over 13 i minus the fraction 32 over 13 j close parenthesis")[$bold(w) = ( display(frac(56, 13)) bold(i) + display(frac(84, 13)) bold(j) ) + ( display(frac(48, 13)) bold(i) − display(frac(32, 13)) bold(j) )$] + #figure(figph[vectors], alt: "vectors", caption: none) #math.equation(block: true, alt: "w equals minus 3 i plus 7 j ,")[$bold(w) = − 3 bold(i) + 7 bold(j) ,$] #math.equation(block: true, alt: "v equals 4 i plus 2 j")[$" " " " bold(v) = 4 bold(i) + 2 bold(j)$] #figure(figph[8x8 grid], alt: "8x8 grid", caption: none) #math.equation(block: true, alt: "w equals 6 i minus 2 j ,")[$bold(w) = 6 bold(i) − 2 bold(j) ,$] #math.equation(block: true, alt: "v equals i minus j")[$" " " " bold(v) = bold(i) − bold(j)$] #figure(figph[8x8 grid], alt: "8x8 grid", caption: none) + #math.equation(block: false, alt: "w equals open parenthesis 4 i minus 4 j close parenthesis plus open parenthesis 2 i plus 2 j close parenthesis")[$bold(w) = ( 4 bold(i) − 4 bold(j) ) + ( 2 bold(i) + 2 bold(j) )$] + #figure(figph[vectors], alt: "vectors", caption: none) #math.equation(block: true, alt: "w equals minus 5 i plus 3 j ,")[$bold(w) = − 5 bold(i) + 3 bold(j) ,$] #math.equation(block: true, alt: "v equals minus i minus 3 j")[$" " " " bold(v) = − bold(i) − 3 bold(j)$] #figure(figph[8x8 grid], alt: "8x8 grid", caption: none) For Problems 11–18, compute the dot product #math.equation(block: false, alt: "u times v")[$bold(u) ⋅ bold(v)$]. #math.equation(block: true, alt: "u equals 3 i plus 7 j ,")[$bold(u) = 3 bold(i) + 7 bold(j) ,$] #math.equation(block: true, alt: "v equals minus 2 i plus 4 j")[$" " " " bold(v) = − 2 bold(i) + 4 bold(j)$] #math.equation(block: true, alt: "22")[$22$] #math.equation(block: true, alt: "u equals minus 1.3 i plus 5.6 j ,")[$bold(u) = − 1.3 bold(i) + 5.6 bold(j) ,$] #math.equation(block: true, alt: "v equals 3 i minus 5 j")[$" " " " bold(v) = 3 bold(i) − 5 bold(j)$] #math.equation(block: true, alt: "u equals 3 i minus 4 j ,")[$bold(u) = 3 bold(i) − 4 bold(j) ,$] #math.equation(block: true, alt: "v equals 20 i plus 15 j")[$" " " " bold(v) = 20 bold(i) + 15 bold(j)$] #math.equation(block: true, alt: "0")[$0$] #math.equation(block: true, alt: "u equals 2 i plus j ,")[$bold(u) = 2 bold(i) + bold(j) ,$] #math.equation(block: true, alt: "v equals 6 i plus 3 j")[$" " " " bold(v) = 6 bold(i) + 3 bold(j)$] #math.equation(block: false, alt: "u")[$bold(u)$] has magnitude 3 and direction #math.equation(block: false, alt: "27 degrees")[$27^(∘)$], and #math.equation(block: false, alt: "v")[$bold(v)$] has magnitude 8 and direction #math.equation(block: false, alt: "minus 33 degrees")[$− 33^(∘)$]. #math.equation(block: true, alt: "12")[$12$] #math.equation(block: false, alt: "u")[$bold(u)$] has magnitude #math.equation(block: false, alt: "the square root of 7")[$sqrt(7)$] and direction #math.equation(block: false, alt: "112 degrees")[$112^(∘)$], and #math.equation(block: false, alt: "v")[$bold(v)$] has magnitude #math.equation(block: false, alt: "the square root of 14")[$sqrt(14)$] and direction #math.equation(block: false, alt: "157 degrees")[$157^(∘)$]. #figure(figph[vectors], alt: "vectors", caption: none) #math.equation(block: true, alt: "minus 318.2")[$− 318.2$] #figure(figph[vectors], alt: "vectors", caption: none) For Problems 19–22, decide whether the pair of vectors is orthogonal. #math.equation(block: false, alt: "2 i plus 3 j")[$2 bold(i) + 3 bold(j) " "$] and #math.equation(block: false, alt: "minus 3 i minus 2 j")[$− 3 bold(i) − 2 bold(j)$] not orthogonal #math.equation(block: false, alt: "minus 5 i plus 7 j")[$− 5 bold(i) + 7 bold(j) " "$] and #math.equation(block: false, alt: "7 i plus 5 j")[$" " 7 bold(i) + 5 bold(j)$] #math.equation(block: false, alt: "4 i plus 6 j")[$4 bold(i) + 6 bold(j) " "$] and #math.equation(block: false, alt: "minus 15 i plus 10 j")[$− 15 bold(i) + 10 bold(j)$] orthogonal #math.equation(block: false, alt: "3 i minus 4 j")[$3 bold(i) − 4 bold(j) " "$] and #math.equation(block: false, alt: "minus 3 i plus 4 j")[$− 3 bold(i) + 4 bold(j)$] For Problems 23–26, find the angle between the vectors. #math.equation(block: false, alt: "3 i plus 5 j")[$3 bold(i) + 5 bold(j) " "$] and #math.equation(block: false, alt: "2 i plus 4 j")[$" " 2 bold(i) + 4 bold(j)$] #math.equation(block: true, alt: "4.4 degrees")[$4.4^(∘)$] #math.equation(block: false, alt: "i minus 2 j")[$bold(i) − 2 bold(j) " "$] and #math.equation(block: false, alt: "minus 2 i minus 3 j")[$− 2 bold(i) − 3 bold(j)$] #math.equation(block: false, alt: "4 i minus 8 j")[$4 bold(i) − 8 bold(j) " "$] and #math.equation(block: false, alt: "6 i plus 4 j")[$" " 6 bold(i) + 4 bold(j)$] #math.equation(block: true, alt: "97.1 degrees")[$97.1^(∘)$] #math.equation(block: false, alt: "minus 6 i plus 8 j")[$− 6 bold(i) + 8 bold(j) " "$] and #math.equation(block: false, alt: "18 i minus 24 j")[$" " 18 bold(i) − 24 bold(j)$] For Problems 27–30, find a value of #math.equation(block: false, alt: "k")[$k$] so that #math.equation(block: false, alt: "v")[$bold(v)$] is orthogonal to #math.equation(block: false, alt: "w")[$bold(w)$]. #math.equation(block: true, alt: "w equals 8 i minus 3 j ,")[$bold(w) = 8 bold(i) − 3 bold(j) ,$] #math.equation(block: true, alt: "v equals 3 i plus k j")[$" " bold(v) = 3 bold(i) + k bold(j)$] #math.equation(block: true, alt: "8")[$8$] #math.equation(block: true, alt: "w equals 2 i plus 7 j ,")[$bold(w) = 2 bold(i) + 7 bold(j) ,$] #math.equation(block: true, alt: "v equals k i plus 4 j")[$" " bold(v) = k bold(i) + 4 bold(j)$] #math.equation(block: true, alt: "w equals minus 2 i minus 5 j ,")[$bold(w) = − 2 bold(i) − 5 bold(j) ,$] #math.equation(block: true, alt: "v equals k i plus 4 j")[$" " bold(v) = k bold(i) + 4 bold(j)$] #math.equation(block: true, alt: "minus 10")[$− 10$] #math.equation(block: true, alt: "w equals 5 i plus 3 j ,")[$bold(w) = 5 bold(i) + 3 bold(j) ,$] #math.equation(block: true, alt: "v equals minus 2 i plus k j")[$" " bold(v) = − 2 bold(i) + k bold(j)$] For Problems 31–36, evaluate the expression for the vectors #math.equation(block: true, alt: "u equals 2 i plus 5 j , v equals minus 3 i plus 4 j , w equals 3 i minus 2 j")[$bold(u) = 2 bold(i) + 5 bold(j) , " " " " bold(v) = − 3 bold(i) + 4 bold(j) , " " " " bold(w) = 3 bold(i) − 2 bold(j)$] #math.equation(block: true, alt: "w times open parenthesis u plus v close parenthesis")[$bold(w) ⋅ ( bold(u) + bold(v) )$] #math.equation(block: true, alt: "minus 21")[$− 21$] #math.equation(block: true, alt: "w times u plus w times v")[$bold(w) ⋅ bold(u) + bold(w) ⋅ bold(v)$] #math.equation(block: true, alt: "open parenthesis u times v close parenthesis w")[$( bold(u) ⋅ bold(v) ) bold(w)$] #math.equation(block: true, alt: "42 i minus 28 j")[$42 bold(i) − 28 bold(j)$] #math.equation(block: true, alt: "open parenthesis u times v close parenthesis open parenthesis u times w close parenthesis")[$( bold(u) ⋅ bold(v) ) ( bold(u) ⋅ bold(w) )$] #math.equation(block: true, alt: "open parenthesis u plus v close parenthesis times open parenthesis u minus v close parenthesis")[$( bold(u) + bold(v) ) ⋅ ( bold(u) − bold(v) )$] #math.equation(block: true, alt: "4")[$4$] #math.equation(block: true, alt: "the fraction w times v over w times w w")[$display(frac(bold(w) ⋅ bold(v), bold(w) ⋅ bold(w))) bold(w)$] Gary pulls a loaded wagon along a flat road. The handle of the wagon makes an angle of #math.equation(block: false, alt: "50 degrees")[$50^(∘)$] to the horizontal. If Gary pulls with a force of 60 pounds, find the component of the force in the direction of motion. 38.57 lbs Wassily is trying to topple a statue by pulling on a rope tied to the statue's upraised arm. The rope is making a #math.equation(block: false, alt: "35 degrees")[$35^(∘)$] angle from horizontal. If Wassily is pulling on the rope with a force of 250 pounds, find the component of the force in the horizontal direction. An SUV weighing 6200 pounds is parked on a hill with slope #math.equation(block: false, alt: "12 degrees")[$12^(∘)$]. Find the force needed to keep the SUV from rolling down the hill. 1289 lbs Steve's boat is headed due north, and the sail points at an angle of #math.equation(block: false, alt: "15 degrees")[$15^(∘)$] east of north. The wind is blowing in the direction #math.equation(block: false, alt: "60 degrees")[$60^(∘)$] west of south, but because of the difference in air pressure between the front and back surfaces of the sail, the boat experiences a force of 400 pounds in the direction the sail is facing. Find the component of the force in the direction of the boat's motion. + Find unit vectors #math.equation(block: false, alt: "u")[$bold(u)$] and #math.equation(block: false, alt: "v")[$bold(v)$] in the directions of #math.equation(block: false, alt: "i plus j")[$bold(i) + bold(j)$] and #math.equation(block: false, alt: "i minus j")[$bold(i) − bold(j)$]. + Show that #math.equation(block: false, alt: "u")[$bold(u)$] and #math.equation(block: false, alt: "v")[$bold(v)$] are orthogonal. + Find the components of #math.equation(block: false, alt: "w equals 3 i plus 8 j")[$bold(w) = 3 bold(i) + 8 bold(j)$] in the directions of #math.equation(block: false, alt: "u")[$bold(u)$] and #math.equation(block: false, alt: "v")[$bold(v)$]. + Sketch the vectors #math.equation(block: false, alt: "u , v")[$bold(u) , " " bold(v)$] and #math.equation(block: false, alt: "w")[$bold(w)$], and show the components of #math.equation(block: false, alt: "w")[$bold(w)$]. + #math.equation(block: false, alt: "the fraction 1 over the square root of 2 i plus the fraction 1 over the square root of 2 j and the fraction minus 1 over the square root of 2 i plus the fraction 1 over the square root of 2 j")[$display(frac(1, sqrt(2))) bold(i) + display(frac(1, sqrt(2))) bold(j) " " " " "and" " " " " display(frac(− 1, sqrt(2))) bold(i) + display(frac(1, sqrt(2))) bold(j)$] + #math.equation(block: false, alt: "u times v equals 0")[$bold(u) ⋅ bold(v) = 0$] + #math.equation(block: false, alt: "the fraction 11 over the square root of 2")[$display(frac(11, sqrt(2)))$] and #math.equation(block: false, alt: "the fraction 5 over the square root of 2")[$display(frac(5, sqrt(2)))$] + #figure(figph[orthogonal vectors], alt: "orthogonal vectors", caption: none) + Find unit vectors #math.equation(block: false, alt: "u")[$bold(u)$] and #math.equation(block: false, alt: "v")[$bold(v)$] in the directions of #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] and #math.equation(block: false, alt: "120 degrees")[$120^(∘)$]. + Show that #math.equation(block: false, alt: "u")[$bold(u)$] and #math.equation(block: false, alt: "v")[$bold(v)$] are orthogonal. + Find the components of #math.equation(block: false, alt: "w equals minus 4 i plus 4 j")[$bold(w) = − 4 bold(i) + 4 bold(j)$] in the directions of #math.equation(block: false, alt: "u")[$bold(u)$] and #math.equation(block: false, alt: "v")[$bold(v)$]. + Sketch the vectors #math.equation(block: false, alt: "u , v")[$bold(u) , " " bold(v)$] and #math.equation(block: false, alt: "w")[$bold(w)$], and show the components of #math.equation(block: false, alt: "w")[$bold(w)$]. For Problems 43–48, let #math.equation(block: false, alt: "u equals a i plus b j")[$bold(u) = a bold(i) + b bold(j)$] and #math.equation(block: false, alt: "v equals c i plus d j")[$bold(v) = c bold(i) + d bold(j)$]. Show that #math.equation(block: false, alt: "v times v equals ∥ v ∥ squared")[$bold(v) ⋅ bold(v) = ∥ bold(v) ∥^(2)$]. #math.equation(block: true, alt: "v times v equals c squared plus d squared")[$bold(v) ⋅ bold(v) = c^(2) + d^(2)$] If #math.equation(block: false, alt: "∥ u ∥ equals 1")[$∥ bold(u) ∥ = 1$], show that #math.equation(block: false, alt: "comp sub u v equals u times v")[$"comp"_(bold(u)) bold(v) = bold(u) ⋅ bold(v)$]. Show that #math.equation(block: false, alt: "k u times v equals k open parenthesis u times v close parenthesis equals u times k v")[$k bold(u) ⋅ bold(v) = k ( bold(u) ⋅ bold(v) ) = bold(u) ⋅ k bold(v)$]. #math.equation(block: true, alt: "k u times v equals k a c plus k b d equals k open parenthesis a c plus b d close parenthesis equals open parenthesis a k c plus b k d close parenthesis")[$k bold(u) ⋅ bold(v) = k a c + k b d = k ( a c + b d ) = ( a k c + b k d )$] Prove the distributive law: #math.equation(block: false, alt: "u times open parenthesis v plus w close parenthesis equals u times v plus u times w")[$bold(u) ⋅ ( bold(v) + bold(w) ) = bold(u) ⋅ bold(v) + bold(u) ⋅ bold(w)$]. Show that #math.equation(block: false, alt: "open parenthesis u minus v close parenthesis times open parenthesis u plus v close parenthesis equals ∥ u ∥ squared minus ∥ v ∥ squared")[$( bold(u) − bold(v) ) ⋅ ( bold(u) + bold(v) ) = ∥ bold(u) ∥^(2) − ∥ bold(v) ∥^(2)$]. #math.equation(block: true, alt: "open parenthesis u minus v close parenthesis times open parenthesis u plus v close parenthesis, equals open parenthesis a minus c close parenthesis open parenthesis a plus c close parenthesis plus open parenthesis b minus d close parenthesis open parenthesis b plus d close parenthesis; equals open parenthesis a squared plus b squared close parenthesis minus open parenthesis c squared plus d squared close parenthesis")[$( bold(u) − bold(v) ) ⋅ ( bold(u) + bold(v) ) & = ( a − c ) ( a + c ) + ( b − d ) ( b + d ) \ & = ( a^(2) + b^(2) ) − ( c^(2) + d^(2) )$] If #math.equation(block: false, alt: "∥ u ∥ equals ∥ v ∥")[$∥ bold(u) ∥ = ∥ bold(v) ∥$], show that #math.equation(block: false, alt: "u plus v")[$bold(u) + bold(v)$] is perpendicular to #math.equation(block: false, alt: "u minus v")[$bold(u) − bold(v)$]. Show that the component of #math.equation(block: false, alt: "v equals a i plus b j")[$bold(v) = a bold(i) + b bold(j)$] in the direction of #math.equation(block: false, alt: "i")[$bold(i)$] is #math.equation(block: false, alt: "a")[$a$], and the component of #math.equation(block: false, alt: "v")[$bold(v)$] in the direction of #math.equation(block: false, alt: "j")[$bold(j)$] is #math.equation(block: false, alt: "b")[$b$]. #math.equation(block: false, alt: "the fraction a times 1 plus b times 0 over 1 equals a")[$display(frac(a ⋅ 1 + b ⋅ 0, 1)) = a$] and #math.equation(block: false, alt: "the fraction a times 0 plus b times 1 over 1 equals b")[$display(frac(a ⋅ 0 + b ⋅ 1, 1)) = b$] Show that the dot product #math.equation(block: false, alt: "u times v")[$bold(u) ⋅ bold(v)$] gives the length of #math.equation(block: false, alt: "v")[$bold(v)$] times the component of #math.equation(block: false, alt: "u")[$bold(u)$] in the direction of #math.equation(block: false, alt: "v")[$bold(v)$]. + Start from the geometric definition #math.equation(block: false, alt: "v times w equals ∥ v ∥ ∥ w ∥ cos ⁡ open parenthesis θ close parenthesis")[$bold(v) ⋅ bold(w) = ∥ bold(v) ∥ ∥ bold(w) ∥ cos ( θ )$] and show that #math.equation(block: false, alt: "i times i equals 1 , j times j equals 1")[$bold(i) ⋅ bold(i) = 1 , " " " " bold(j) ⋅ bold(j) = 1$] and #math.equation(block: false, alt: "i times j equals 0")[$bold(i) ⋅ bold(j) = 0$]. + Use part (a) and Problems 45 and 46 to derive the coordinate definition of #math.equation(block: false, alt: "v times w")[$bold(v) ⋅ bold(w)$]. + Both #math.equation(block: false, alt: "i times i equals 1")[$bold(i) ⋅ bold(i) = 1$] and #math.equation(block: false, alt: "j times j equals 1")[$bold(j) ⋅ bold(j) = 1$] because #math.equation(block: false, alt: "1 times 1 cos ⁡ 0 equals 1")[$1 ⋅ 1 cos 0 = 1$]; #math.equation(block: false, alt: "i times j equals 1 times 1 cos ⁡ 90 degrees equals 0")[$bold(i) ⋅ bold(j) = 1 ⋅ 1 cos 90^(∘) = 0$] + #math.equation(block: false, alt: "open parenthesis a i plus b j close parenthesis times open parenthesis c i plus d j close parenthesis equals a c open parenthesis 1 close parenthesis plus a d open parenthesis 0 close parenthesis plus b c open parenthesis 0 close parenthesis plus b d open parenthesis 1 close parenthesis equals a c plus b d")[$( a bold(i) + b bold(j) ) ⋅ ( c bold(i) + d bold(j) ) = a c ( 1 ) + a d ( 0 ) + b c ( 0 ) + b d ( 1 ) = a c + b d$] + Show that #math.equation(block: false, alt: "open parenthesis v plus w close parenthesis times open parenthesis v plus w close parenthesis equals ∥ v ∥ squared plus ∥ w ∥ squared plus 2 open parenthesis v times w close parenthesis")[$( bold(v) + bold(w) ) ⋅ ( bold(v) + bold(w) ) = ∥ bold(v) ∥^(2) + ∥ bold(w) ∥^(2) + 2 ( bold(v) ⋅ bold(w) )$] + Use part (a) to prove the triangle inequality:#math.equation(block: true, alt: "∥ v plus w ∥ less than or equal to ∥ v ∥ plus ∥ w ∥")[$∥ bold(v) + bold(w) ∥ ≤ ∥ bold(v) ∥ + ∥ bold(w) ∥$] + Use the dot product to show that #math.equation(block: false, alt: "∥ u minus v ∥ squared equals ∥ u ∥ squared plus ∥ v ∥ squared minus 2 ∥ u ∥ ∥ v ∥ cos ⁡ open parenthesis θ close parenthesis")[$∥ bold(u) − bold(v) ∥^(2) = ∥ bold(u) ∥^(2) + ∥ bold(v) ∥^(2) − 2 ∥ bold(u) ∥ ∥ bold(v) ∥ cos ( θ )$]. + Use the figure at right to explain why part (a) proves the law of cosines. #figure(figph[vectors], alt: "vectors", caption: none) + #math.equation(block: false, alt: "∥ u minus v ∥ squared equals u times u minus 2 u times v plus v times v equals ∥ u ∥ squared plus ∥ v ∥ squared minus 2 ∥ u ∥ ∥ v ∥ cos ⁡ θ")[$∥ bold(u) − bold(v) ∥^(2) = bold(u) ⋅ bold(u) − 2 bold(u) ⋅ bold(v) + bold(v) ⋅ bold(v) = ∥ bold(u) ∥^(2) + ∥ bold(v) ∥^(2) − 2 ∥ bold(u) ∥ ∥ bold(v) ∥ cos θ$] + Let #math.equation(block: false, alt: "a equals ∥ u ∥ , b equals ∥ v ∥ , c equals ∥ u minus v ∥")[$a = ∥ bold(u) ∥ , " " b = ∥ bold(v) ∥ , " " c = ∥ bold(u) − bold(v) ∥$], and #math.equation(block: false, alt: "C equals θ")[$C = θ$] + If #math.equation(block: false, alt: "u")[$bold(u)$] is a unit vector, and the angle between #math.equation(block: false, alt: "u")[$bold(u)$] and #math.equation(block: false, alt: "i")[$bold(i)$] is #math.equation(block: false, alt: "α")[$α$], show that #math.equation(block: false, alt: "u equals cos ⁡ open parenthesis α close parenthesis i plus sin ⁡ open parenthesis α close parenthesis j")[$bold(u) = cos ( α ) bold(i) + sin ( α ) bold(j)$]. + Suppose #math.equation(block: false, alt: "u")[$bold(u)$] and #math.equation(block: false, alt: "v")[$bold(v)$] are unit vectors, as shown in the figure at right. Use the dot product #math.equation(block: false, alt: "u times v")[$bold(u) ⋅ bold(v)$] to prove that #math.equation(block: true, alt: "cos ⁡ open parenthesis β minus α close parenthesis equals cos ⁡ open parenthesis β close parenthesis cos ⁡ open parenthesis α close parenthesis plus sin ⁡ open parenthesis β close parenthesis sin ⁡ open parenthesis α close parenthesis")[$cos ( β − α ) = cos ( β ) cos ( α ) + sin ( β ) sin ( α )$] #figure(figph[vectors], alt: "vectors", caption: none)