Question

Let $\overrightarrow{a}$ be a unit vector and $\overrightarrow{b}$ be a non-zero vector not parallel to $\overrightarrow{a}$. If two sides of a triangle are represented by the vectors $\sqrt{3}(\overrightarrow{a}\times \overrightarrow{b})$ and $\overrightarrow{b}-(\overrightarrow{a}.\overrightarrow{b})\overrightarrow{a}$, then the angles of the triangle are

• A. ${30}^0,{90}^0,{60}^0$
• B. ${45}^0,{45}^0,{90}^0$
• C. ${60}^0,{60}^0,{60}^0$
• D. $noneofthese$

Answer 1

The correct option is A ${30}^0,{90}^0,{60}^0$

Observe that $\overline{b}-(\overline{a}.\overline{b})\overline{a}=(\overline{a}.\overline{a})\overline{b}-(\overline{a}.\overline{b})\overline{a}$ since $\overline{a}$ is a unit vector.

And thus, it becomes equal to $(\overline{a}\times \overline{b})\times \overline{a}$.

Now, the two sides become perpendicular to each other!

Also, $(\overline{a}\times \overline{b})$ being perpendicular to $\overline{a}$ and $\overline{a}$ being a unit vector, the ratios of the lengths of two sides becomes $\sqrt{3}:1.$

Hence the rest two angles can thus be found out to be ${60}^o$ and ${30}^o.$