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Chapter 10 · Vector Algebra

Exercise 10.2 · Q9: Unit vector along a⃗+b⃗\vec a+\vec b for 2i^−j^+2k^2\hat i-\hat j+2\hat k⁠, −i^+j^−k^-\hat i+\hat j-\hat k

Exercise 10.2Sum of vectors, unit vector · Sections 10.4, 10.5 and 10.5.1
The question

For given vectors, a⃗=2i^−j^+2k^\vec a = 2\hat i - \hat j + 2\hat k and b⃗=−i^+j^−k^\vec b = -\hat i + \hat j - \hat k⁠, find the unit vector in the direction of the vector a⃗+b⃗\vec a + \vec b⁠.

Exam answer a⃗+b⃗=(2−1)i^+(−1+1)j^+(2−1)k^=i^+0j^+k^=i^+k^\begin{aligned} \vec a + \vec b &= (2 - 1)\hat i + (-1 + 1)\hat j \\ &\quad + (2 - 1)\hat k \\ &= \hat i + 0\hat j + \hat k = \hat i + \hat k \end{aligned} ∣a⃗+b⃗∣=12+02+12=2\begin{aligned} |\vec a + \vec b| &= \sqrt{1^2 + 0^2 + 1^2} \\ &= \sqrt2 \end{aligned}

The unit vector in the direction of a⃗+b⃗\vec a + \vec b is

a⃗+b⃗∣a⃗+b⃗∣=i^+k^2\frac{\vec a + \vec b}{|\vec a + \vec b|} = \frac{\hat i + \hat k}{\sqrt2}

∴\therefore the unit vector is 12i^+12k^\dfrac{1}{\sqrt2}\hat i + \dfrac{1}{\sqrt2}\hat k⁠.

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

The one idea this problem is built on

Two short jobs in a row. First add the vectors component by component to get the single vector a⃗+b⃗\vec a + \vec b⁠. Then divide that vector by its own length to get the unit vector in its direction.

Given, to find, and the plan
Given
a⃗=2i^−j^+2k^\vec a = 2\hat i - \hat j + 2\hat k and b⃗=−i^+j^−k^\vec b = -\hat i + \hat j - \hat k⁠.
To find
The unit vector in the direction of a⃗+b⃗\vec a + \vec b⁠.
Plan
(1) Add the components. (2) Find ∣a⃗+b⃗∣|\vec a + \vec b|⁠. (3) Divide. (4) Check the length, and see why the sum is so short.

Basics you need

1Sum in component form

a⃗+b⃗=(a1+b1)i^+(a2+b2)j^\vec a + \vec b = (a_1 + b_1)\hat i + (a_2 + b_2)\hat j +(a3+b3)k^+ (a_3 + b_3)\hat k (Section 10.5.1, (i)). Lesson: adding arrows, head to tail

2Magnitude 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

3The unit vector

For c⃗≠0⃗\vec c \ne \vec 0⁠, c^=1∣c⃗∣c⃗\hat c = \frac{1}{|\vec c|}\vec c is the unit vector in the direction of c⃗\vec c (Section 10.5). Lesson: components

Solution

  1. Add the two vectors

    a⃗+b⃗=(2−1)i^+(−1+1)j^+(2−1)k^=i^+0j^+k^=i^+k^.\begin{aligned} \vec a + \vec b &= (2 - 1)\hat i + (-1 + 1)\hat j \\ &\quad + (2 - 1)\hat k \\ &= \hat i + 0\hat j + \hat k = \hat i + \hat k. \end{aligned}
    Vectors in component form are added by adding the matching components. The components of b⃗\vec b are −1-1⁠, 11 and −1-1 (a term with no number has component 1, with its sign). The j^\hat j parts cancel, so the sum has no j^\hat j part at all.
    a⃗+b⃗\vec a + \vec b by components (Section 10.5.1, (i)). Lesson: adding arrows, head to tail

  2. Find the length of the sum

    ∣a⃗+b⃗∣=12+02+12=2.|\vec a + \vec b| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt2.
    The missing j^\hat j part counts as 0. The length is not 0, so a⃗+b⃗\vec a + \vec b is not the zero vector and has a direction.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  3. Divide by the length

    a⃗+b⃗∣a⃗+b⃗∣=i^+k^2=12i^+12k^.\begin{aligned} \frac{\vec a + \vec b}{|\vec a + \vec b|} &= \frac{\hat i + \hat k}{\sqrt2} \\ &= \frac{1}{\sqrt2}\hat i + \frac{1}{\sqrt2}\hat k. \end{aligned}

    This is the unit vector in the direction of a⃗+b⃗\vec a + \vec b.

    Dividing by the positive number 2\sqrt2 keeps the direction and makes the length 1. Rationalised, each component is 22\frac{\sqrt2}{2}⁠. The j^\hat j component stays 0.
    c^=1∣c⃗∣c⃗\hat c = \frac{1}{|\vec c|}\vec c (Section 10.5). Lesson: components

  4. Check, and see why the sum is short

    (12)2+02+(12)2=12+12=1.\left(\tfrac{1}{\sqrt2}\right)^2 + 0^2 + \left(\tfrac{1}{\sqrt2}\right)^2 = \tfrac12 + \tfrac12 = 1.

    So the length is 1. Notice that ∣a⃗∣=4+1+4=3|\vec a| = \sqrt{4 + 1 + 4} = 3 and ∣b⃗∣=1+1+1=3|\vec b| = \sqrt{1 + 1 + 1} = \sqrt3⁠, yet ∣a⃗+b⃗∣=2|\vec a + \vec b| = \sqrt2 is smaller than both: b⃗\vec b points almost back along a⃗\vec a⁠.

    O A C a = 2î − ĵ + 2k̂, length 3 b, length √3 a + b = î + k̂ unit vector along a + b: (î + k̂)/√2
    The triangle law drawn in the plane of a⃗\vec a and b⃗\vec b⁠, true to scale with one scale: OA→=a⃗\overrightarrow{OA} = \vec a (length 3), AC→=b⃗\overrightarrow{AC} = \vec b (length 3\sqrt3⁠) and OC→=a⃗+b⃗\overrightarrow{OC} = \vec a + \vec b (length 2\sqrt2⁠). The short thick arrow along OC is the unit vector, length 1.
    By the triangle law, a⃗+b⃗\vec a + \vec b is the third side OC of a triangle with sides of lengths 3 and 3\sqrt3⁠. When the second arrow turns back towards the start, the third side is short. The answer is the direction of OC, whatever the lengths of a⃗\vec a and b⃗\vec b⁠.
    Triangle law of vector addition (Section 10.4, Fig 10.8). Compare Example 8, the same task for another pair. Lesson: adding arrows, head to tail
Result

a⃗+b⃗=i^+k^\vec a + \vec b = \hat i + \hat k⁠, and the unit vector in its direction is 12i^+12k^\frac{1}{\sqrt2}\hat i + \frac{1}{\sqrt2}\hat k⁠.

(Textbook answer: 12i^+12k^\frac{1}{\sqrt2}\hat i + \frac{1}{\sqrt2}\hat k⁠.)

Check it with numbers

∣a⃗+b⃗∣=2≈1.414214|\vec a + \vec b| = \sqrt2 \approx 1.414214⁠, while ∣a⃗∣+∣b⃗∣=3+3≈4.732051|\vec a| + |\vec b| = 3 + \sqrt3 \approx 4.732051⁠: the sum is much shorter than the two lengths added.

Each nonzero component of the unit vector is 12≈0.707107\frac{1}{\sqrt2} \approx 0.707107⁠, so the unit vector makes 45∘45^\circ with OX and OZ and 90∘90^\circ with OY (its j^\hat j component is 0, so it lies in the xz-plane).

Common mistakes
  1. Sign slips in b⃗\vec b⁠. b⃗=−i^+j^−k^\vec b = -\hat i + \hat j - \hat k has components −1,1,−1-1, 1, -1⁠; reading them as 1,1,11, 1, 1 gives 3i^+3k^3\hat i + 3\hat k⁠, a different direction.
  2. Adding the lengths. ∣a⃗+b⃗∣|\vec a + \vec b| is not ∣a⃗∣+∣b⃗∣=3+3|\vec a| + |\vec b| = 3 + \sqrt3⁠; find the sum first, then its length.
  3. Dropping k^\hat k or keeping a stray j^\hat j⁠. The sum is i^+0j^+k^\hat i + 0\hat j + \hat k⁠: the j^\hat j part is exactly 0.
What to take away
  1. Add component by component, then find the length of the result.
  2. The unit vector along a⃗+b⃗\vec a + \vec b is a⃗+b⃗∣a⃗+b⃗∣\frac{\vec a + \vec b}{|\vec a + \vec b|}⁠.
  3. ∣a⃗+b⃗∣|\vec a + \vec b| can be much smaller than ∣a⃗∣+∣b⃗∣|\vec a| + |\vec b| when the vectors point in nearly opposite directions.

Every fact used, and where it comes from

FactUsed inSource
a⃗+b⃗\vec a + \vec b by componentsStep 1Section 10.5.1, (i) (Lesson: adding arrows)
∣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 (Basics: use Pythagoras twice)
c^=1∣c⃗∣c⃗\hat c = \frac{1}{|\vec c|}\vec cStep 3Section 10.5
Triangle law of vector additionStep 4Section 10.4, Fig 10.8
The answerResultNCERT answer key, Exercise 10.2 Q9 (Lesson: components, lengths and unit vectors)
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