The sine of the angle between vectors a = 2 i - 6 j - 3 k and b = 4 i + 3 j - k is

Correct Answer is : 526{5}{{ {26} }}265

Correct Answer is : 526{5}{{ {26} }}265

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UPSC NDA I 2019 Mathematics Solved Paper

Question : 69 of 120

Marks: +1, -0

\vec a = 2\hat i - 6\hat j - 3\hat k

\vec b = 4\hat i + 3\hat j - \hat k

Solution:

Concept:

The sine of the angle

\theta

between two vectors

\vec a

and

\vec b

is found using the identity

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

, where

\cos\theta = \frac{\vec a \cdot \vec b}{|\vec a|\,|\vec b|}

Explanation:

Given

\vec a = 2\hat i - 6\hat j - 3\hat k

and

\vec b = 4\hat i + 3\hat j - \hat k

Compute the dot product:

\vec a \cdot \vec b = (2)(4) + (-6)(3) + (-3)(-1) = 8 - 18 + 3 = -7

Compute magnitudes:

|\vec a| = \sqrt{2^2 + (-6)^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7

;

|\vec b| = \sqrt{4^2 + 3^2 + (-1)^2} = \sqrt{16 + 9 + 1} = \sqrt{26}

Then

\cos\theta = \frac{-7}{7 \times \sqrt{26}} = -\frac{1}{\sqrt{26}}

Thus

\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{1}{26} = \frac{25}{26}

Taking the positive square root (since

\theta

is between

0^\circ

and

180^\circ

\sin\theta = \frac{5}{\sqrt{26}}

Answer:

\frac{5}{\sqrt{26}}

(Option B).

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