Question

If $a=(2,1,–1),b=(1,–1,0),c=(5,–1,1),$ then unit vector parallel to $a+b–c$ but in opposite direction is

• A. $\frac{1}{3}(2\hat{i}-\hat{j}+2\hat{k})$
• B. $\frac{1}{2}(2\hat{i}-\hat{j}+2\hat{k})$
• C. $\frac{1}{3}(2\hat{i}-\hat{j}-2\hat{k})$
• D. None of these

Answer 1

The correct option is A

$\frac{1}{3}(2\hat{i}-\hat{j}+2\hat{k})$

Find the unit vector parallel to $\mathit{a}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}\mathbf{}\mathbf{–}\mathbf{}\mathit{c}$ :

Given that $a=(2,1,–1),b=(1,–1,0),c=(5,–1,1),$

$a+b-c=\hat{i}(2+1-5)+\hat{j}(1-1+1)+\hat{k}(-1+0-1)=-2i+j-2k$

Unit vector of $a+b-c=\frac{(a+b-c)}{|a+b-c|}$

$$\begin{gathered} =\frac{(-2\hat{i}+\hat{j}-2\hat{k})}{\sqrt{4+1+4}} \\ =\frac{(-2\hat{i}+\hat{j}-2\hat{k})}{3} \end{gathered}$$

Therefore, the vector opposite to this unit vector is;

$\frac{1}{3}(2\hat{i}-\hat{j}+2\hat{k})$

Hence the correct option is A.