Question

If $\overrightarrow{A}=i-j,\overrightarrow{B}=i+j+k$ are two vectors and $\overrightarrow{C}$ is another vectors such that $\overrightarrow{A}\times \overrightarrow{C}+\overrightarrow{B}=\overrightarrow{0}$ and $\overrightarrow{A}.\overrightarrow{C}=0$, then $|\overrightarrow{C}{|}^2$ =

• A. $9$
• B. $8$
• C. $\frac{19}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is D $\frac{3}{2}$

$\overrightarrow{A}=\hat{ı}-\hat{ȷ}\overrightarrow{B}=\hat{i}+\hat{ȷ}+\hat{k}$

1et $\overrightarrow{c}=x\hat{ı}+y\hat{ȷ}+z\hat{k}$

Given that $\overrightarrow{A}\cdot \overrightarrow{C}=0$

$\therefore (\hat{ı}-\hat{ȷ})\cdot (x\hat{ı}+y\hat{ȷ}+z\hat{k})=0$

$\Rightarrow (x-y)=0$

$\Rightarrow x=y$

Given that $\overrightarrow{A}\times \overrightarrow{C}+\overrightarrow{B}=\overrightarrow{0}$

$\overrightarrow{A}\times \overrightarrow{C}=\left|\begin{array}{ccc}\hat{ı}& \hat{ȷ}& \hat{k}\\ 1& -1& 0\\ x& y& z\end{array}\right|$

$=\hat{i}(-z)-\hat{j}\left(z\right)+\hat{k}(y+x)$

$=-z\hat{i}-z\hat{ȷ}+2x\hat{k}$

$(\overrightarrow{A}\times \overrightarrow{C})+\overrightarrow{B}=(-z\hat{ı}-z\hat{ȷ}+2x\hat{k})+(\hat{ı}+\hat{ȷ}+\hat{k})$

$=(1-z)\hat{ı}+(1-z)\hat{ȷ}+(2x+1)\hat{k}$

$(\overrightarrow{A}\times \overrightarrow{C})+\overrightarrow{B}=\overrightarrow{0}$

$\therefore (1-z)\hat{ı}+(1-z)\hat{ȷ}+(2x+1)\hat{k}=0\hat{ı}+0\hat{ȷ}+0\hat{k}$

Comparing both sides

$z=1x=-\frac{1}{2}=y$

$\therefore \overrightarrow{C}=-\frac{1}{2}\hat{ı}-\frac{1}{2}\hat{ȷ}+\hat{k}$

$|\overrightarrow{C}|=\sqrt{{(-\frac{1}{2})}^2+{(-\frac{1}{2})}^2+(1{)}^2}=\sqrt{\frac{1}{2}+1}$

$|\overrightarrow{C}|=\sqrt{\frac{3}{2}}$

$|\overrightarrow{C}{|}^2=\frac{3}{2}$