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. $({90}^o-{\cos }^{-1}\sqrt{\frac{5}{11}})$
• D. $({180}^o-{\cos }^{-1}\sqrt{\frac{5}{11}})$

Answer 1

Answer: (A)

${\cos }^{-1}\sqrt{\frac{5}{11}}$

$\overrightarrow{AB}=3\hat{i}+\hat{j}+\hat{k}$
$\overrightarrow{AC}=\hat{i}+2\hat{j}+\hat{k}$
$\overrightarrow{CB}=\overrightarrow{AB}-\overrightarrow{AC}$
$2\hat{i}-\hat{j}$
$\angle$ABC is angle between $\overrightarrow{AB}$ and $\overrightarrow{CB}$
$\overrightarrow{AB}\cdot \overrightarrow{CB}=|\overrightarrow{AB}||\overrightarrow{CB}|\cos \theta$
$\Rightarrow 5=|\sqrt{11}||\sqrt{5}|\cos \theta$
$\cos \theta =\frac{\sqrt{5}}{\sqrt{11}}$
$\theta ={\cos }^{-1}\left(\frac{\sqrt{5}}{\sqrt{11}}\right)$.