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UPSC NDA I 2023 Mathematics Solved Paper

Question Numbers: 67-68

Consider the following for the next two (02) items that follow :

Let a vector

\vec{a}=4\hat{i}-8\hat{j}+\hat{k}

make angles α, β, γ with the positive directions of x, y, z axes respectively.

Question : 68 of 120

Marks: +1, -0

What is cos2β + cos2γ equal to ?

Solution:

Concept:

For a vector

\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}

, the direction cosines are

\cos\alpha = \frac{x}{|\vec{a}|}

\cos\beta = \frac{y}{|\vec{a}|}

\cos\gamma = \frac{z}{|\vec{a}|}

, and they satisfy

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

Explanation:

Given

\vec{a} = 4\hat{i} - 8\hat{j} + \hat{k}

and

\cos\alpha = \frac{4}{9}

We need

\cos2\beta + \cos2\gamma

Using the identity

\cos2\theta = 2\cos^2\theta - 1

, we write:

\cos2\beta + \cos2\gamma = (2\cos^2\beta - 1) + (2\cos^2\gamma - 1) = 2(\cos^2\beta + \cos^2\gamma) - 2

From the direction cosine identity:

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

Thus,

\cos2\beta + \cos2\gamma = 2(1 - \cos^2\alpha) - 2 = 2 - 2\cos^2\alpha - 2 = -2\cos^2\alpha

Substitute

\cos\alpha = \frac{4}{9}

\cos^2\alpha = \frac{16}{81}

, so the expression becomes

-2 \times \frac{16}{81} = -\frac{32}{81}

Answer:

\boxed{-\frac{32}{81}}

(Option A).

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