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CBSE Sample Paper 2022-23 · Section A · 1 mark

Q8: Projection of 3i^−j^−2k^3\hat i - \hat j - 2\hat k on i^+2j^−3k^\hat i + 2\hat j - 3\hat k

Sample 2022-23Ch 10 · Vector Algebra
The question

The scalar projection of the vector 3i^−j^−2k^3\hat i - \hat j - 2\hat k on the vector i^+2j^−3k^\hat i + 2\hat j - 3\hat k is

(a) 714\frac{7}{\sqrt{14}}

(b) 714\frac{7}{14}

(c) 613\frac{6}{13}

(d) 72\frac{7}{2}

Exam answer

Let a⃗=3i^−j^−2k^\vec a = 3\hat i - \hat j - 2\hat k and b⃗=i^+2j^−3k^\vec b = \hat i + 2\hat j - 3\hat k⁠.

a⃗⋅b⃗=3−2+6=7∣b⃗∣=1+4+9=14\begin{aligned} \vec a \cdot \vec b &= 3 - 2 + 6 = 7 \\ |\vec b| &= \sqrt{1 + 4 + 9} = \sqrt{14} \end{aligned}
Why?

The dot product multiplies matching components and adds:

(3)(1)+(−1)(2)+(−2)(−3)=3−2+6=1+6=7\begin{aligned} &(3)(1) + (-1)(2) \\ &\quad + (-2)(-3) \\ &= 3 - 2 + 6 \\ &= 1 + 6 = 7 \end{aligned}

Watch the last one: (−2)(−3)=+6(-2)(-3) = +6⁠.

The magnitude is the square root of the sum of the squares of the components, 12+22+(−3)2=141^2 + 2^2 + (-3)^2 = 14⁠.

(1)2+(2)2+(−3)2=1+4+9=14\begin{aligned} &(1)^2 + (2)^2 + (-3)^2 \\ &= 1 + 4 + 9 \\ &= 14 \end{aligned}
Projection of a⃗ on b⃗=a⃗⋅b⃗∣b⃗∣=714\begin{aligned} &\text{Projection of } \vec a \text{ on } \vec b \\ &= \frac{\vec a \cdot \vec b}{|\vec b|} = \frac{7}{\sqrt{14}} \end{aligned}
Why?

The scalar projection of a⃗\vec a on b⃗\vec b is the length of the shadow of a⃗\vec a along b⃗\vec b⁠:

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

It divides by the length of b⃗\vec b⁠, the vector we project on.

Watch out: dividing by ∣b⃗∣2=14|\vec b|^2 = 14 instead gives 714\frac{7}{14}⁠, option (b), which is not the scalar projection.

Answer: (a) 714\dfrac{7}{\sqrt{14}}

Where this comes from

Chapter 10, Vector Algebra: Scalar projection of a vector on a vector.

Question type: Projection

How to spot it: click to see

Type 6. Projection: the shadow of one vector on another

Spot it: "find the projection of the vector a⃗\vec a on the vector b⃗\vec b".

Here: Project 3i^−j^−2k^3\hat i - \hat j - 2\hat k on i^+2j^−3k^\hat i + 2\hat j - 3\hat k⁠: the dot product is 7, then divide by the length of the second vector, 14\sqrt{14}⁠.

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