Question

Two sides of a triangle are formed by the vectors $\overrightarrow{a}=3\hat{i}+6\hat{j}-2\hat{k}$ and $\overrightarrow{b}=4\hat{i}-\hat{j}+3\hat{k}$. The angles of the triangle are
1) ${\cos }^{-1}\frac{7}{\sqrt{75}}$

2 ${\cos }^{-1}\left(\sqrt{\frac{26}{75}}\right)$

3) ${\cos }^{-1}\frac{3}{\sqrt{15}}$

4 ${\cos }^{-1}\frac{2}{3}$

• A. $1$ & $4$
• B. $2$ & $3$
• C. $3$ & $4$
• D. $1$ & $2$

Answer 1

The correct option is D $1$ & $2$

$\overline{a}.\overline{b}=0$, so angle between them is $\pi /2$

Therefore the third side is $\overline{c}=\overline{a}+\overline{b}=7i+5j+k$

now we can find other two angles

$\alpha =\arccos \frac{\overline{c}.\overline{a}}{|a|.|c|}=\arccos \frac{7}{\sqrt{75}}$

$\beta =\arccos \frac{\overline{c}.\overline{b}}{|b|.|c|}=\arccos \frac{\sqrt{26}}{\sqrt{75}}$