Q215 / 269 · NDA 2025 (Phase 1) · Mathematics

A line makes angles $\alpha$, $\beta$ and $\gamma$ with the positive directions of the coordinate axes. If $\vec{a} = (\sin^2\alpha)\hat{i} + (\sin^2\beta)\hat{j} + (\sin^2\gamma)\hat{k}$ and $\vec{b} = \hat{i} + \hat{j} + \hat{k}$, then what is $\vec{a} \cdot \vec{b}$ equal to?

  • A
    $-2$
  • B
    $-1$
  • C
    $1$
  • Correct answer
    D
    $2$

Solution

Let $\cos\alpha$, $\cos\beta$, and $\cos\gamma$ be the direction cosines of the given line.

We know that the sum of the squares of the direction cosines is $1$:

$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$

The given vectors are:

$\vec{a} = (\sin^2\alpha)\hat{i} + (\sin^2\beta)\hat{j} + (\sin^2\gamma)\hat{k}$

$\vec{b} = \hat{i} + \hat{j} + \hat{k}$

The dot product $\vec{a} \cdot \vec{b}$ is given by:

$\vec{a} \cdot \vec{b} = (\sin^2\alpha)(1) + (\sin^2\beta)(1) + (\sin^2\gamma)(1)$

$\vec{a} \cdot \vec{b} = \sin^2\alpha + \sin^2\beta + \sin^2\gamma$

Using the identity $\sin^2\theta = 1 - \cos^2\theta$, we get:

$\vec{a} \cdot \vec{b} = (1 - \cos^2\alpha) + (1 - \cos^2\beta) + (1 - \cos^2\gamma)$

$\vec{a} \cdot \vec{b} = 3 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma)$

Substituting $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$:

$\vec{a} \cdot \vec{b} = 3 - 1 = 2$

Answer: $2$

Hence, the correct answer is option: 'D'.

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