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Chapter 10 · Vector Algebra

Exercise 10.2 · Q10: Vector of magnitude 8 along 5i^−j^+2k^5\hat i-\hat j+2\hat k

Exercise 10.2Unit vector, vector of given magnitude · Sections 10.5 and 10.5.1
The question

Find a vector in the direction of vector 5i^−j^+2k^5\hat i - \hat j + 2\hat k which has magnitude 8 units.

Exam answer

Let a⃗=5i^−j^+2k^\vec a = 5\hat i - \hat j + 2\hat k⁠.

∣a⃗∣=52+(−1)2+22=30\begin{aligned} |\vec a| &= \sqrt{5^2 + (-1)^2 + 2^2} \\ &= \sqrt{30} \end{aligned} ∴ a^=130(5i^−j^+2k^)\therefore\ \hat a = \frac{1}{\sqrt{30}}(5\hat i - \hat j + 2\hat k)

The vector of magnitude 8 in the direction of a⃗\vec a is 8a^8\hat a⁠:

8a^=830(5i^−j^+2k^)8\hat a = \frac{8}{\sqrt{30}}(5\hat i - \hat j + 2\hat k)

∴\therefore the required vector is 4030i^−830j^+1630k^\dfrac{40}{\sqrt{30}}\hat i - \dfrac{8}{\sqrt{30}}\hat j + \dfrac{16}{\sqrt{30}}\hat k⁠.

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

The one idea this problem is built on

The unit vector a^\hat a carries only the direction (its length is 1). Multiplying it by 8 keeps that direction and makes the length 8. So: find a^\hat a⁠, then take 8a^8\hat a⁠.

Given, to find, and the plan
Given
The direction of a⃗=5i^−j^+2k^\vec a = 5\hat i - \hat j + 2\hat k⁠, and the magnitude 8.
To find
The vector of magnitude 8 in the direction of a⃗\vec a⁠.
Plan
(1) Find ∣a⃗∣|\vec a|⁠. (2) Form the unit vector a^\hat a⁠. (3) Multiply by 8 and distribute. (4) Check that the length is 8 and the direction is that of a⃗\vec a⁠.

Basics you need

1Magnitude from components

∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

2The unit vector

For a⃗≠0⃗\vec a \ne \vec 0⁠, a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec a has length 1 and the direction of a⃗\vec a (Section 10.5). Lesson: six kinds of vectors worth naming

3Scaling a vector

For λ>0\lambda > 0⁠, λa⃗\lambda\vec a has the direction of a⃗\vec a and magnitude λ∣a⃗∣\lambda|\vec a|⁠; in components, each component is multiplied by λ\lambda (Sections 10.5 and 10.5.1, (iv)). Lesson: components

Solution

  1. Find the magnitude of the given vector

    ∣a⃗∣=52+(−1)2+22=25+1+4=30.\begin{aligned} |\vec a| &= \sqrt{5^2 + (-1)^2 + 2^2} \\ &= \sqrt{25 + 1 + 4} = \sqrt{30}. \end{aligned}
    The j^\hat j component is −1-1⁠, and (−1)2=1(-1)^2 = 1⁠: a minus sign never makes a length smaller. Since 30≠0\sqrt{30} \ne 0⁠, the vector has a direction.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  2. The unit vector in that direction

    a^=130(5i^−j^+2k^).\hat a = \frac{1}{\sqrt{30}}(5\hat i - \hat j + 2\hat k).
    Dividing by the length gives the vector of length 1 pointing the same way. It holds the direction and nothing else, which is what we need before choosing the length.
    a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec a (Section 10.5). Lesson: six kinds of vectors worth naming

  3. Stretch it to length 8

    8a^=830(5i^−j^+2k^)=4030i^−830j^+1630k^.\begin{aligned} 8\hat a &= \frac{8}{\sqrt{30}}(5\hat i - \hat j + 2\hat k) \\ &= \frac{40}{\sqrt{30}}\hat i - \frac{8}{\sqrt{30}}\hat j + \frac{16}{\sqrt{30}}\hat k. \end{aligned}

    This is the required vector.

    8>08 > 0⁠, so 8a^8\hat a has the direction of a^\hat a⁠, the direction of a⃗\vec a⁠, and its magnitude is 8×1=88 \times 1 = 8⁠. Each component is multiplied by 830\frac{8}{\sqrt{30}}⁠: 5⋅8=405 \cdot 8 = 40⁠, (−1)⋅8=−8(-1) \cdot 8 = -8⁠, 2⋅8=162 \cdot 8 = 16⁠. Rationalised, the same vector is 4303i^−43015j^+83015k^\frac{4\sqrt{30}}{3}\hat i - \frac{4\sqrt{30}}{15}\hat j + \frac{8\sqrt{30}}{15}\hat k⁠.
    ∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a|⁠, and λa⃗\lambda\vec a by components (Sections 10.5 and 10.5.1, (iv)). Lesson: components

  4. Check the length and the direction

    ∣8a^∣2=402+82+16230=1600+64+25630=192030=64.\begin{aligned} |8\hat a|^2 &= \frac{40^2 + 8^2 + 16^2}{30} \\ &= \frac{1600 + 64 + 256}{30} \\ &= \frac{1920}{30} = 64. \end{aligned}

    So ∣8a^∣=8|8\hat a| = 8⁠. And 8a^=830a⃗8\hat a = \frac{8}{\sqrt{30}}\vec a is a positive multiple of a⃗\vec a⁠, so the direction is right.

    â, length 1 a, length √30 ≈ 5.48 8â, length 8 same direction, lengths 1 : √30 : 8
    Three arrows in the direction of a⃗\vec a⁠, drawn from the same starting line with one scale: a^\hat a (length 1), a⃗\vec a itself (length 30\sqrt{30}⁠) and the answer 8a^8\hat a (length 8). They differ only in length.
    The two conditions of the question are "in the direction of 5i^−j^+2k^5\hat i - \hat j + 2\hat k" and "magnitude 8". The first holds because the answer is a positive multiple of a⃗\vec a⁠; the second is the calculation above.
    λa⃗\lambda\vec a has the same direction as a⃗\vec a when λ>0\lambda > 0 (Section 10.5). This is the method of Example 7, with 8 in place of 7.
Result

The required vector is 8a^=830(5i^−j^+2k^)8\hat a = \frac{8}{\sqrt{30}}(5\hat i - \hat j + 2\hat k) =4030i^−830j^+1630k^= \frac{40}{\sqrt{30}}\hat i - \frac{8}{\sqrt{30}}\hat j + \frac{16}{\sqrt{30}}\hat k⁠.

(Textbook answer: 4030i^−830j^+1630k^\frac{40}{\sqrt{30}}\hat i - \frac{8}{\sqrt{30}}\hat j + \frac{16}{\sqrt{30}}\hat k⁠.)

Check it with numbers

30≈5.477226\sqrt{30} \approx 5.477226⁠, so the scale factor is 830≈1.460593\frac{8}{\sqrt{30}} \approx 1.460593⁠. The answer is about 7.302967i^−1.460593j^+2.921187k^7.302967\hat i - 1.460593\hat j + 2.921187\hat k⁠.

The ratios of the components are exactly those of a⃗\vec a⁠: 40:(−8):16=5:(−1):240 : (-8) : 16 = 5 : (-1) : 2⁠, each multiplied by 8.

Common mistakes
  1. Answering 8a⃗8\vec a⁠. 8a⃗=40i^−8j^+16k^8\vec a = 40\hat i - 8\hat j + 16\hat k has magnitude 8308\sqrt{30}⁠, not 8. Multiply the unit vector, not the vector.
  2. Losing the minus sign. The j^\hat j component is −830-\frac{8}{\sqrt{30}}⁠; with ++ the direction changes.
  3. Taking −8a^-8\hat a⁠. It has magnitude 8 but points the opposite way, so it is not "in the direction of" a⃗\vec a⁠.
What to take away
  1. Vector of magnitude mm in the direction of a⃗\vec a⁠: ma^=m∣a⃗∣a⃗m\hat a = \frac{m}{|\vec a|}\vec a⁠.
  2. The unit vector holds the direction; the number in front sets the length.
  3. Check: the squares of the components add to m2m^2⁠.

Every fact used, and where it comes from

FactUsed inSource
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Steps 1, 4Section 10.5.1 (Basics: use Pythagoras twice)
a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec aStep 2Section 10.5
∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a|⁠; same direction for λ>0\lambda > 0Steps 3, 4Section 10.5
λa⃗\lambda\vec a by componentsStep 3Section 10.5.1, (iv) (Lesson: components)
The answerResultNCERT answer key, Exercise 10.2 Q10 (Lesson: components, lengths and unit vectors)
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