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Chapter 10 · Vector Algebra

Exercise 10.3 · Q4: Projection of i^+3j^+7k^\hat i+3\hat j+7\hat k on 7i^−j^+8k^7\hat i-\hat j+8\hat k

Exercise 10.3Projection of a vector on a vector · Section 10.6.2
The question

Find the projection of the vector i^+3j^+7k^\hat i + 3\hat j + 7\hat k on the vector 7i^−j^+8k^7\hat i - \hat j + 8\hat k⁠.

Exam answer

Let a⃗=i^+3j^+7k^\vec a = \hat i + 3\hat j + 7\hat k and b⃗=7i^−j^+8k^\vec b = 7\hat i - \hat j + 8\hat k⁠. The projection of a⃗\vec a on b⃗\vec b is a⃗⋅b⃗∣b⃗∣\frac{\vec a\cdot\vec b}{|\vec b|}⁠.

a⃗⋅b⃗=(1)(7)+(3)(−1)+(7)(8)=7−3+56=60\begin{aligned} \vec a\cdot\vec b &= (1)(7) + (3)(-1) + (7)(8) \\ &= 7 - 3 + 56 = 60 \end{aligned} ∣b⃗∣=72+(−1)2+82=49+1+64=114\begin{aligned} |\vec b| &= \sqrt{7^2 + (-1)^2 + 8^2} \\ &= \sqrt{49 + 1 + 64} = \sqrt{114} \end{aligned} Projection=a⃗⋅b⃗∣b⃗∣=60114\text{Projection} = \frac{\vec a\cdot\vec b}{|\vec b|} = \frac{60}{\sqrt{114}}

∴\therefore the projection of a⃗\vec a on b⃗\vec b is 60114\dfrac{60}{\sqrt{114}}⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The projection of a⃗\vec a on b⃗\vec b is the signed length of the shadow of a⃗\vec a on the line of b⃗\vec b⁠: ∣a⃗∣cos⁡θ|\vec a|\cos\theta⁠. Since a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta⁠, it equals

∣a⃗∣cos⁡θ=a⃗⋅b⃗∣b⃗∣=a⃗⋅b^.|\vec a|\cos\theta = \frac{\vec a\cdot\vec b}{|\vec b|} = \vec a\cdot\hat b.

So we need only the scalar product and the length of the vector we project on.

Given, to find, and the plan
Given
a⃗=i^+3j^+7k^\vec a = \hat i + 3\hat j + 7\hat k⁠, b⃗=7i^−j^+8k^\vec b = 7\hat i - \hat j + 8\hat k⁠.
To find
The projection of a⃗\vec a on b⃗\vec b⁠.
Plan
(1) Compute a⃗⋅b⃗\vec a\cdot\vec b⁠. (2) Compute ∣b⃗∣|\vec b|⁠. (3) Divide (and rationalise if wanted). (4) Check that the shadow is not longer than a⃗\vec a⁠.

Basics you need

1Projection on a vector

The projection of a⃗\vec a on b⃗\vec b is a⃗⋅b^=a⃗⋅b⃗∣b⃗∣=1∣b⃗∣(a⃗⋅b⃗)\vec a\cdot\hat b = \vec a\cdot\frac{\vec b}{|\vec b|} = \frac{1}{|\vec b|}(\vec a\cdot\vec b) (Section 10.6.2, Observation 2, Fig 10.20). Lesson: the dot product

2Scalar product in components

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3 (Section 10.6.1). Lesson: Examples 13–21

3Cosine as a shadow

A segment of length rr at angle θ\theta to a line casts a shadow of signed length rcos⁡θr\cos\theta on it, and ∣cos⁡θ∣≤1|\cos\theta| \le 1⁠. Basics: cos θ is a shadow

Solution

  1. The scalar product

    a⃗⋅b⃗=(1)(7)+(3)(−1)+(7)(8)=7−3+56=60.\begin{aligned} \vec a\cdot\vec b &= (1)(7) + (3)(-1) + (7)(8) \\ &= 7 - 3 + 56 = 60. \end{aligned}
    Multiply matching components and add. The j^\hat j-component of b⃗\vec b is −1-1 (the term −j^-\hat j⁠), so the middle product is 3×(−1)=−33 \times (-1) = -3⁠.
    a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3 (Section 10.6.1). Lesson: the dot product

  2. The length of b⃗\vec b

    ∣b⃗∣=72+(−1)2+82=49+1+64=114.\begin{aligned} |\vec b| &= \sqrt{7^2 + (-1)^2 + 8^2} \\ &= \sqrt{49 + 1 + 64} = \sqrt{114}. \end{aligned}
    We project on b⃗\vec b⁠, so we divide by the length of b⃗\vec b⁠, the vector that fixes the line. The length of a⃗\vec a is not used here. 114=2×3×19114 = 2 \times 3 \times 19 has no square factor, so 114\sqrt{114} does not simplify.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  3. Divide

    a⃗⋅b⃗∣b⃗∣=60114=60114114=1011419.\frac{\vec a\cdot\vec b}{|\vec b|} = \frac{60}{\sqrt{114}} = \frac{60\sqrt{114}}{114} = \frac{10\sqrt{114}}{19}.

    So the projection of a⃗\vec a on b⃗\vec b is 60114\frac{60}{\sqrt{114}}, about 5.62.

    l O a, length √59 b, length √114 projection 60/√114 ≈ 5.62 cos θ ≈ 0.73 θ ≈ 42.98°
    a⃗\vec a and b⃗\vec b drawn in their own plane, true to scale. The perpendicular from the tip of a⃗\vec a to the line ll of b⃗\vec b marks the shadow of a⃗\vec a⁠: its length is the projection 60114≈5.62\frac{60}{\sqrt{114}} \approx 5.62⁠, less than ∣a⃗∣=59≈7.68|\vec a| = \sqrt{59} \approx 7.68⁠. It is positive because a⃗⋅b⃗>0\vec a\cdot\vec b > 0⁠: the angle between the vectors is acute.
    60114\frac{60}{\sqrt{114}} is the book's form. Multiplying top and bottom by 114\sqrt{114} gives 60114114\frac{60\sqrt{114}}{114}⁠, and cancelling 6 gives 1011419\frac{10\sqrt{114}}{19}⁠; both forms are the same number. The answer is positive, so the shadow points the same way as b⃗\vec b⁠.
    Projection of a⃗\vec a on b⃗\vec b (Section 10.6.2, Observation 2). Lesson: Examples 13–21

  4. Check against the length of a⃗\vec a

    ∣a⃗∣=1+9+49=59|\vec a| = \sqrt{1 + 9 + 49} = \sqrt{59}⁠. The projection is ∣a⃗∣cos⁡θ|\vec a|\cos\theta and ∣cos⁡θ∣≤1|\cos\theta| \le 1⁠, so it can never be longer than 59\sqrt{59}⁠. Indeed (60114)2=3600114=60019\left(\frac{60}{\sqrt{114}}\right)^2 = \frac{3600}{114} = \frac{600}{19}⁠, which is less than 59.

    This guards against the commonest slip, dividing by ∣a⃗∣|\vec a| instead of ∣b⃗∣|\vec b|⁠: that would give 6059\frac{60}{\sqrt{59}}⁠, the projection of b⃗\vec b on a⃗\vec a⁠, which must be at most ∣b⃗∣=114|\vec b| = \sqrt{114}⁠.
    ∣cos⁡θ∣≤1|\cos\theta| \le 1⁠. Basics: cos θ is a shadow
Result

The projection of i^+3j^+7k^\hat i + 3\hat j + 7\hat k on 7i^−j^+8k^7\hat i - \hat j + 8\hat k is 60114=1011419\frac{60}{\sqrt{114}} = \frac{10\sqrt{114}}{19}⁠.

(Textbook answer: 60114\frac{60}{\sqrt{114}}⁠.)

Check it with numbers

114≈10.677078\sqrt{114} \approx 10.677078⁠, so 60114≈5.619515\frac{60}{\sqrt{114}} \approx 5.619515⁠, and 1011419≈5.619515\frac{10\sqrt{114}}{19} \approx 5.619515 (the same number), while ∣a⃗∣=59≈7.681146|\vec a| = \sqrt{59} \approx 7.681146⁠.

The angle: cos⁡θ=6059 114=606726≈0.731599\cos\theta = \frac{60}{\sqrt{59}\,\sqrt{114}} = \frac{60}{\sqrt{6726}} \approx 0.731599⁠, so θ≈42.979∘\theta \approx 42.979^\circ⁠, and 59cos⁡θ=60114\sqrt{59}\cos\theta = \frac{60}{\sqrt{114}} exactly.

The projection vector (the shadow as a vector along b⃗\vec b⁠) is a⃗⋅b⃗∣b⃗∣2b⃗=60114(7i^−j^+8k^)\frac{\vec a\cdot\vec b}{|\vec b|^2}\vec b = \frac{60}{114}(7\hat i - \hat j + 8\hat k) =1019(7i^−j^+8k^)= \frac{10}{19}(7\hat i - \hat j + 8\hat k)⁠, whose length is 1011419\frac{10\sqrt{114}}{19} again.

Common mistakes
  1. Dividing by ∣a⃗∣|\vec a|⁠. 6059\frac{60}{\sqrt{59}} is the projection of b⃗\vec b on a⃗\vec a⁠; projecting on b⃗\vec b divides by ∣b⃗∣|\vec b|⁠.
  2. Reading −j^-\hat j as +j^+\hat j⁠. Then a⃗⋅b⃗=7+3+56=66\vec a\cdot\vec b = 7 + 3 + 56 = 66⁠, which is wrong; the j^\hat j-component of b⃗\vec b is −1-1⁠.
  3. Forgetting to divide at all. a⃗⋅b⃗=60\vec a\cdot\vec b = 60 is ∣b⃗∣|\vec b| times the projection, not the projection.
What to take away
  1. Projection of a⃗\vec a on b⃗\vec b⁠: a⃗⋅b⃗∣b⃗∣\frac{\vec a\cdot\vec b}{|\vec b|}⁠, dividing by the length of the vector projected on.
  2. It is the signed length of the shadow, ∣a⃗∣cos⁡θ|\vec a|\cos\theta⁠, so its size is at most ∣a⃗∣|\vec a|⁠.
  3. 60114\frac{60}{\sqrt{114}} and 1011419\frac{10\sqrt{114}}{19} are the same answer; rationalising is optional.

Every fact used, and where it comes from

FactUsed inSource
a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3Step 1Section 10.6.1 (Lesson: the dot product)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Steps 2, 4Section 10.5.1
Projection of a⃗\vec a on b⃗\vec b is a⃗⋅b^=1∣b⃗∣(a⃗⋅b⃗)\vec a\cdot\hat b = \frac{1}{|\vec b|}(\vec a\cdot\vec b)Step 3Section 10.6.2, Observation 2, Fig 10.20
∣cos⁡θ∣≤1|\cos\theta| \le 1Step 4Trigonometry (Basics: cos θ is a shadow)
Projection vectorCheckSection 10.6.2
The answerResultNCERT answer key, Exercise 10.3 Q4 (Lesson: the dot product)
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