Question

In a triangle $ABC$, the sides $AB$ and $AC$ are represented by the vectors $3 \hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}$ respectively. Calculate the angle $\angle ABC$

• A. $\cos^{-1} \sqrt{\frac{5}{11}}$
• B. $\cos^{-1} \sqrt{\frac{6}{11}}$
• C. $\left( 90^{\circ} - \cos^{-1} \sqrt{\frac{5}{11}} \right)$
• D. $\left( 180^{\circ} - \cos^{-1} \sqrt{\frac{5}{11}} \right)$

Answer 1

The Correct Option is A

$ A B =3 \hat{ i }+\hat{ j }+\hat{ k } \text { and } A C =\hat{ i }+2 \hat{ j }+\hat{ k } $
$\therefore B A =-(3 \hat{ i }+\hat{ j }+\hat{ k })$
$\angle A B C$ is angle between $B A$ and $B C$

$ B C = A C + B A $
$= A C - A B (\because B A =- A B ) $
$= \hat{i} + 2\hat{j} + \hat{k} - (3\hat{i} + \hat{j} + \hat{k}) = -2\hat{i} + \hat{j}$
$\therefore BA \cdot BC = |BA||BC| \cos\,\theta$
$-(3\hat{i} + \hat{j} + \hat{k}) \cdot (-2 \hat{i} + \hat{j})$
$= (\sqrt{3^2 + 1^2 + 1^2)} \cdot (\sqrt{2^2 + 1^2}) \cos\,\theta$
$\Rightarrow 6 - 1 = \sqrt{11 \times 5} \cdot \cos\,theta$
$\therefore \cos \,\theta = \frac{5}{\sqrt{55}} = \sqrt{\frac{5}{11}}$
$ = \theta = \cos^{-1} = \sqrt{\frac{5}{11}}$