Aspire Faculty ID #11945 · Topic: NIMCET 2025 · Just now
NIMCET 2025

The length of the projection of $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ on $\vec{b} = -2\hat{i} + \hat{j} + 2\hat{k}$, is equal to:

Solution

$\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ $\vec{b} = -2\hat{i} + \hat{j} + 2\hat{k}$ 
The length of projection of $\vec{a}$ on $\vec{b}$ is: 
$\displaystyle \text{Proj length} = \frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$ 
Compute dot product: $\vec{a}\cdot\vec{b} = (2)(-2) + (3)(1) + (1)(2)$ $= -4 + 3 + 2 = 1$ 
Compute $|\vec{b}|$: $|\vec{b}| = \sqrt{(-2)^2 + 1^2 + 2^2}$ $= 3$ 
Thus projection length: $\displaystyle \frac{1}{3}$
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