Question

Angle between the vectors $\overrightarrow{A}=-2\hat{i}-2\hat{j}+2\hat{k}$ and $\overrightarrow{B}=x\hat{i}+\hat{j}+(x+1)\hat{k}$ is

• A. ${90}^0$
• B. $0^0$
• C. an acute angle
• D. an obtuse angle

Answer 1

Answer: (A)

${90}^0$

$\begin{array}{c}\overrightarrow{A}=-2\hat{i}-2\hat{j}+2\hat{k}\\ \overrightarrow{B}=x\hat{i}+\hat{j}+(x+1)\hat{k}\end{array}$

$|\overrightarrow{A}|=\sqrt{4+4+4}=\sqrt{12}$

$|\overrightarrow{B}|=\sqrt{x^2+1+(x+1{)}^2}$

$=\sqrt{x^2+1+x^2+1+2x^2}$

$=\sqrt{2x^2+2x+2}$

$\overrightarrow{A}\cdot \overrightarrow{B}=-2x-2+2(x+1)$

$=0$

Let $\theta$ be the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$

then,

$\cos \theta =\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|}=\frac{0}{\sqrt{12}\sqrt{2x^2+2x+2}}$

$\cos \theta =0$

$\theta ={90}^{\circ }$