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Chapter 10 · Vector Algebra

Miscellaneous · Q7: Unit vector parallel to 2a⃗−b⃗+3c⃗2\vec a-\vec b+3\vec c

Miscellaneous ExerciseLinear combinations, unit vectors · Sections 10.5 and 10.5.1
The question

If a⃗=i^+j^+k^\vec a = \hat i + \hat j + \hat k⁠, b⃗=2i^−j^+3k^\vec b = 2\hat i - \hat j + 3\hat k and c⃗=i^−2j^+k^\vec c = \hat i - 2\hat j + \hat k⁠, find a unit vector parallel to the vector 2a⃗−b⃗+3c⃗2\vec a - \vec b + 3\vec c⁠.

Exam answer

Let r⃗=2a⃗−b⃗+3c⃗\vec r = 2\vec a - \vec b + 3\vec c⁠. Then

r⃗=2(i^+j^+k^)−(2i^−j^+3k^)+3(i^−2j^+k^)=(2−2+3)i^+(2+1−6)j^+(2−3+3)k^=3i^−3j^+2k^\begin{aligned} \vec r &= 2(\hat i + \hat j + \hat k) \\ &\quad - (2\hat i - \hat j + 3\hat k) \\ &\quad + 3(\hat i - 2\hat j + \hat k) \\ &= (2 - 2 + 3)\hat i + (2 + 1 - 6)\hat j \\ &\quad + (2 - 3 + 3)\hat k \\ &= 3\hat i - 3\hat j + 2\hat k \end{aligned} ∣r⃗∣=32+(−3)2+22=9+9+4=22\begin{aligned} |\vec r| &= \sqrt{3^2 + (-3)^2 + 2^2} \\ &= \sqrt{9 + 9 + 4} = \sqrt{22} \end{aligned}

The unit vector parallel to r⃗\vec r is

r^=r⃗∣r⃗∣=122(3i^−3j^+2k^)\hat r = \frac{\vec r}{|\vec r|} = \frac{1}{\sqrt{22}}(3\hat i - 3\hat j + 2\hat k)

∴ r^=322i^−322j^+222k^\therefore\ \hat r = \dfrac{3}{\sqrt{22}}\hat i - \dfrac{3}{\sqrt{22}}\hat j + \dfrac{2}{\sqrt{22}}\hat k⁠.

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

The one idea this problem is built on

First turn 2a⃗−b⃗+3c⃗2\vec a - \vec b + 3\vec c into a single vector in components; the scalar multiples and the sum work component by component. Then divide that vector by its own length to get the unit vector in its direction.

Given, to find, and the plan
Given
a⃗=i^+j^+k^\vec a = \hat i + \hat j + \hat k⁠, b⃗=2i^−j^+3k^\vec b = 2\hat i - \hat j + 3\hat k⁠, c⃗=i^−2j^+k^\vec c = \hat i - 2\hat j + \hat k⁠.
To find
A unit vector parallel to r⃗=2a⃗−b⃗+3c⃗\vec r = 2\vec a - \vec b + 3\vec c⁠.
Plan
(1) Work out 2a⃗2\vec a⁠, −b⃗-\vec b⁠, 3c⃗3\vec c⁠. (2) Add them. (3) Find ∣r⃗∣|\vec r|⁠. (4) Divide. (5) Check the length is 1.

Basics you need

1Scalar multiples and sums by components

λ(a1i^+a2j^+a3k^)=λa1i^+λa2j^+λa3k^\lambda(a_1\hat i + a_2\hat j + a_3\hat k) = \lambda a_1\hat i + \lambda a_2\hat j + \lambda a_3\hat k⁠, and sums add matching components (Section 10.5.1). Lesson: components

2Length 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

3Unit vector

r^=1∣r⃗∣r⃗\hat r = \frac{1}{|\vec r|}\vec r is the unit vector in the direction of r⃗\vec r (Section 10.5). Lesson: six kinds of vectors

Solution

  1. The three pieces

    2a⃗=2i^+2j^+2k^,−b⃗=−2i^+j^−3k^,3c⃗=3i^−6j^+3k^.\begin{aligned} 2\vec a &= 2\hat i + 2\hat j + 2\hat k, \\ -\vec b &= -2\hat i + \hat j - 3\hat k, \\ 3\vec c &= 3\hat i - 6\hat j + 3\hat k. \end{aligned}
    Each scalar multiplies every component. The minus sign in front of b⃗\vec b is the scalar −1-1⁠, so every sign of b⃗\vec b flips: −(−j^)=+j^-(-\hat j) = +\hat j⁠.
    Scalar multiple by components (Section 10.5.1, (iv)). Lesson: components

  2. Add them

    r⃗=(2−2+3)i^+(2+1−6)j^+(2−3+3)k^=3i^−3j^+2k^.\begin{aligned} \vec r &= (2 - 2 + 3)\hat i + (2 + 1 - 6)\hat j \\ &\quad + (2 - 3 + 3)\hat k \\ &= 3\hat i - 3\hat j + 2\hat k. \end{aligned}
    Collect the i^\hat i parts from the three lines of Step 1, then the j^\hat j parts, then the k^\hat k parts.
    Sum by components (Section 10.5.1, (i)). Lesson: components

  3. The length

    ∣r⃗∣=32+(−3)2+22=22.|\vec r| = \sqrt{3^2 + (-3)^2 + 2^2} = \sqrt{22}.
    9+9+4=229 + 9 + 4 = 22⁠. The square of −3-3 is +9+9⁠, so the minus sign does not reduce the length.
    Magnitude from components (Section 10.5.1). Basics: use Pythagoras twice

  4. Divide by the length

    r^=122(3i^−3j^+2k^)=322i^−322j^+222k^.\begin{aligned} \hat r &= \frac{1}{\sqrt{22}}(3\hat i - 3\hat j + 2\hat k) \\ &= \frac{3}{\sqrt{22}}\hat i - \frac{3}{\sqrt{22}}\hat j + \frac{2}{\sqrt{22}}\hat k. \end{aligned}

    So r^=122(3i^−3j^+2k^)\hat r = \frac{1}{\sqrt{22}}(3\hat i - 3\hat j + 2\hat k).

    Multiplying by the positive number 122\frac{1}{\sqrt{22}} keeps the direction of r⃗\vec r and scales the length 22\sqrt{22} down to 1. A vector in the same direction is parallel to r⃗\vec r⁠.
    Unit vector in a given direction (Section 10.5); compare Exercise 10.2 Q9. Lesson: components, lengths and unit vectors

  5. Check, and the opposite unit vector

    922+922+422=2222=1.\tfrac{9}{22} + \tfrac{9}{22} + \tfrac{4}{22} = \tfrac{22}{22} = 1.

    So ∣r^∣=1|\hat r| = 1⁠. The negative −r^-\hat r is also a unit vector parallel to r⃗\vec r⁠, pointing the other way.

    The squares of the components of a unit vector add to 1. "A unit vector parallel" asks for one; r^\hat r⁠, in the same direction as r⃗\vec r⁠, is the one the answer key gives.
    Collinear vectors (Section 10.5.1, Remarks (i)). Lesson: put it all together
Result

2a⃗−b⃗+3c⃗=3i^−3j^+2k^2\vec a - \vec b + 3\vec c = 3\hat i - 3\hat j + 2\hat k⁠, and the unit vector parallel to it is 322i^−322j^+222k^\frac{3}{\sqrt{22}}\hat i - \frac{3}{\sqrt{22}}\hat j + \frac{2}{\sqrt{22}}\hat k (its negative is the other one).

(Textbook answer: 322i^−322j^+222k^\frac{3}{\sqrt{22}}\hat i - \frac{3}{\sqrt{22}}\hat j + \frac{2}{\sqrt{22}}\hat k⁠.)

Check it with numbers

22≈4.690416\sqrt{22} \approx 4.690416⁠, 322≈0.639602\frac{3}{\sqrt{22}} \approx 0.639602 and 222≈0.426401\frac{2}{\sqrt{22}} \approx 0.426401⁠.

A second route to r⃗\vec r⁠, adding a⃗\vec a twice, c⃗\vec c three times and taking away b⃗\vec b in a different order: a⃗+a⃗=2i^+2j^+2k^\vec a + \vec a = 2\hat i + 2\hat j + 2\hat k⁠; + 3c⃗+\,3\vec c gives 5i^−4j^+5k^5\hat i - 4\hat j + 5\hat k⁠; − b⃗-\,\vec b gives 3i^−3j^+2k^3\hat i - 3\hat j + 2\hat k⁠, the same.

Common mistakes
  1. Forgetting to flip all the signs of b⃗\vec b⁠. −b⃗=−2i^+j^−3k^-\vec b = -2\hat i + \hat j - 3\hat k⁠; keeping −j^-\hat j gives the wrong j^\hat j part −5-5⁠.
  2. Multiplying only the first component. 3c⃗=3i^−6j^+3k^3\vec c = 3\hat i - 6\hat j + 3\hat k⁠, not 3i^−2j^+k^3\hat i - 2\hat j + \hat k⁠.
  3. Stopping at r⃗\vec r⁠. The question asks for a unit vector: divide by 22\sqrt{22}⁠.
What to take away
  1. A combination like 2a⃗−b⃗+3c⃗2\vec a - \vec b + 3\vec c is worked out component by component.
  2. Unit vector along r⃗\vec r⁠: r⃗∣r⃗∣\frac{\vec r}{|\vec r|}⁠.
  3. Check: the squares of a unit vector's components add to 1.

Every fact used, and where it comes from

FactUsed inSource
Scalar multiple and sum by componentsSteps 1, 2Section 10.5.1, (i) and (iv) (Lesson: components)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Steps 3, 5Section 10.5.1 (Basics: use Pythagoras twice)
Unit vector r^=1∣r⃗∣r⃗\hat r = \frac{1}{|\vec r|}\vec rStep 4Section 10.5
Collinear vectorsStep 5Section 10.5.1, Remarks (i)
The answerResultNCERT answer key, Miscellaneous Exercise on Chapter 10, Q7 (Lesson: put it all together)
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