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

• A. ${\cos }^{-1}\frac{7}{\sqrt{75}}$
• B. ${\cos }^{-1}\sqrt{\frac{26}{75}}$
• C. ${\cos }^{-1}\frac{3}{\sqrt{15}}$
• D. ${\cos }^{-1}\frac{2}{3}$

Answer 1

Answer: (A), (B)

The correct options are
A ${\cos }^{-1}\frac{7}{\sqrt{75}}$
B ${\cos }^{-1}\sqrt{\frac{26}{75}}$

Since $\overrightarrow{a}.\overrightarrow{b}=0$, the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac{\pi }{2}$

Now, adding these two vectors, we get

$\overrightarrow{c}=\overrightarrow{a}+\overrightarrow{b}=7\hat{i}+5\hat{j}+\hat{k}$

So, the other two angles of the triangle are given by

$\alpha ={\cos }^{-1}\frac{\overrightarrow{c}.\overrightarrow{a}}{\left|\overrightarrow{c}\right|\left|\overrightarrow{a}\right|}={\cos }^{-1}\frac{\left(7\right)\left(3\right)+\left(5\right)\left(6\right)+\left(1\right)(-2)}{\sqrt{7^2+5^2+1^2}\sqrt{3^2+6^2+{(-2)}^2}}={\cos }^{-1}\frac{7}{\sqrt{75}}$

$\beta ={\cos }^{-1}\frac{\overrightarrow{c}.\overrightarrow{b}}{\left|\overrightarrow{c}\right|\left|\overrightarrow{b}\right|}={\cos }^{-1}\frac{\left(7\right)\left(4\right)+\left(5\right)(-1)+\left(1\right)\left(3\right)}{\sqrt{7^2+5^2+1^2}\sqrt{4^2+{(-1)}^2+3^2}}={\cos }^{-1}\sqrt{\frac{26}{75}}$