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NCERT Exemplar · Q40

Q.Projection vector of a⃗\vec{a} on b⃗\vec{b} is
(A) (a⃗⋅b⃗∣b⃗∣2)b⃗\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}
(B) a⃗⋅b⃗∣b⃗∣\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}
(C) a⃗⋅b⃗∣a⃗∣\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|}
(D) a⃗⋅b⃗∣a⃗∣2b^\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2}\hat{b}

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The projection vector of a⃗\vec{a} on b⃗\vec{b} is the vector component of a⃗\vec{a} along the direction of b⃗\vec{b}. It is given by (a⃗⋅b⃗∣b⃗∣2)b⃗\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}, which is option (A).

The idea of a projection is simple: if you shine a light straight down onto a line, the shadow a vector casts on that line is its projection. For vectors, the projection of a⃗\vec{a} onto b⃗\vec{b} answers: "How much of a⃗\vec{a} points in the direction of b⃗\vec{b}, and what is that vector?"

This is not a scalar — it is a vector itself. It has a magnitude (the length of the shadow) and a direction (the direction of b⃗\vec{b}). So the formula must produce a vector that is parallel to b⃗\vec{b}.

Let’s build it step by step.

  1. Find the scalar component of a⃗\vec{a} along b⃗\vec{b}. The dot product a⃗⋅b⃗\vec{a}\cdot\vec{b} gives ∣a⃗∣∣b⃗∣cos⁡θ|\vec{a}||\vec{b}|\cos\theta, where θ\theta is the angle between them. The quantity ∣a⃗∣cos⁡θ|\vec{a}|\cos\theta is the length of the projection of a⃗\vec{a} onto the line of b⃗\vec{b}. To isolate this, divide the dot product by ∣b⃗∣|\vec{b}|:

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

This is the scalar projection (also called the component of a⃗\vec{a} along b⃗\vec{b}). It tells you how long the shadow is, but not the vector itself.

  1. Turn that scalar into a vector. To get the actual projection vector, we need to multiply this scalar length by a unit vector in the direction of b⃗\vec{b}. The unit vector along b⃗\vec{b} is b^=b⃗∣b⃗∣\hat{b} = \frac{\vec{b}}{|\vec{b}|}. So:

Projection vector of a⃗ on b⃗=(a⃗⋅b⃗∣b⃗∣)b^=(a⃗⋅b⃗∣b⃗∣)b⃗∣b⃗∣=(a⃗⋅b⃗∣b⃗∣2)b⃗.\text{Projection vector of } \vec{a} \text{ on } \vec{b} = \left(\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\right) \hat{b} = \left(\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\right) \frac{\vec{b}}{|\vec{b}|} = \left(\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right) \vec{b}.

  1. Match with the options. Option (A) is exactly (a⃗⋅b⃗∣b⃗∣2)b⃗\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}. Option (B) is the scalar projection (missing the direction). …

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