Question

Two vectors $(x^2-1)\hat{i}+(x+2)\hat{j}+x^2\hat{k}$ and $2\hat{i}-x\hat{j}+3\hat{k}$ are orthogonal

• A. for no real value of $x$
• B. for $x=-1$
• C. for $x=\frac{1}{2}$
• D. for $x=-\frac{1}{2}$ and $x=1$

Answer 1

Answer: (D)

for $x=-\frac{1}{2}$ and $x=1$

Given two vectors are orthogonal if,

For orthogonality dot product should be zero.

$\begin{array}{c}2(x^2-1)-x(x+2)+3x^2=0\\ \\ 2x^2-2-x^2-2x+3x^2=0\\ \\ 4x^2-2x-2=0\\ \\ 2x^2-x-1=0\\ \\ 2x^2+x-2x-1=0\\ \\ x\left(2x+1\right)-1\left(2x+1\right)=0\\ \\ \left(2x+1\right)\left(x-1\right)=0\\ \\ x=1,-\frac{1}{2}\end{array}$

Hence, option $D$ is correct.