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Vector Algebra - Projection of a vector on a line

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The projection of a vector a⃗\vec{a} on a directed line ll (or another vector b⃗\vec{b}) is the scalar value representing the magnitude of the 'shadow' of a⃗\vec{a} along that line.

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If θ\theta is the angle between a⃗\vec{a} and b⃗\vec{b}, the projection of a⃗\vec{a} on b⃗\vec{b} is given by ∣a⃗∣cos⁡θ|\vec{a}| \cos \theta.

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The projection of a⃗\vec{a} on b⃗\vec{b} can also be expressed as the dot product of a⃗\vec{a} with the unit vector b^\hat{b} in the direction of b⃗\vec{b}.

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If the angle θ=0\theta = 0, the projection is ∣a⃗∣|\vec{a}|. If θ=π\theta = \pi, the projection is −∣a⃗∣-|\vec{a}|. If θ=π2\theta = \frac{\pi}{2} or 3π2\frac{3\pi}{2}, the projection is 00.

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Vector projection: While the projection is usually a scalar, the 'vector projection' of a⃗\vec{a} on b⃗\vec{b} is a vector whose magnitude is the scalar projection and whose direction is that of b⃗\vec{b} (or opposite if the projection is negative).

📐Formulae

Projection of a⃗ on b⃗=a⃗⋅b⃗∣b⃗∣\text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

Projection of a⃗ on b⃗=a⃗⋅b^\text{Projection of } \vec{a} \text{ on } \vec{b} = \vec{a} \cdot \hat{b}

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

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3

∣b⃗∣=b12+b22+b32|\vec{b}| = \sqrt{b_1^2 + b_2^2 + b_3^2}

💡Examples

Problem 1:

Find the projection of the vector a⃗=i^+3j^+7k^\vec{a} = \hat{i} + 3\hat{j} + 7\hat{k} on the vector b⃗=7i^−j^+8k^\vec{b} = 7\hat{i} - \hat{j} + 8\hat{k}.

Solution:

  1. Find the dot product a⃗⋅b⃗\vec{a} \cdot \vec{b}: a⃗⋅b⃗=(1)(7)+(3)(−1)+(7)(8)\vec{a} \cdot \vec{b} = (1)(7) + (3)(-1) + (7)(8) a⃗⋅b⃗=7−3+56=60\vec{a} \cdot \vec{b} = 7 - 3 + 56 = 60

  2. Find the magnitude of b⃗\vec{b}: ∣b⃗∣=72+(−1)2+82|\vec{b}| = \sqrt{7^2 + (-1)^2 + 8^2} ∣b⃗∣=49+1+64=114|\vec{b}| = \sqrt{49 + 1 + 64} = \sqrt{114}

  3. Calculate the projection: Projection=a⃗⋅b⃗∣b⃗∣=60114\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{60}{\sqrt{114}}

Explanation:

To find the projection of a⃗\vec{a} on b⃗\vec{b}, we calculate the dot product of the two vectors and divide it by the magnitude of the vector on which the projection is being taken (vector b⃗\vec{b}).

Problem 2:

Find the value of λ\lambda if the projection of a⃗=λi^+j^+4k^\vec{a} = \lambda\hat{i} + \hat{j} + 4\hat{k} on b⃗=2i^+6j^+3k^\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k} is 44 units.

Solution:

The formula for projection is: a⃗⋅b⃗∣b⃗∣=4\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 4

Calculate a⃗⋅b⃗\vec{a} \cdot \vec{b}: a⃗⋅b⃗=(λ)(2)+(1)(6)+(4)(3)=2λ+6+12=2λ+18\vec{a} \cdot \vec{b} = (\lambda)(2) + (1)(6) + (4)(3) = 2\lambda + 6 + 12 = 2\lambda + 18

Calculate ∣b⃗∣|\vec{b}|: ∣b⃗∣=22+62+32=4+36+9=49=7|\vec{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7

Substitute into the projection formula: 2λ+187=4\frac{2\lambda + 18}{7} = 4 2λ+18=282\lambda + 18 = 28 2λ=102\lambda = 10 λ=5\lambda = 5

Explanation:

We use the given scalar projection value in the standard formula to set up an algebraic equation and solve for the unknown parameter λ\lambda.