# The value of

Question:

Let $\vec{a}=\hat{i}-\hat{j}, \vec{b}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c}+\vec{b}=\overrightarrow{0}$ and $\vec{a} \cdot \vec{c}=4$, then $|\vec{c}|^{2}$ is equal to:-

1. $\frac{19}{2}$

2. 8

3. $\frac{17}{2}$

4. 9

Correct Option: 1

Solution:

$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{b}}$

$(\vec{a} \times \vec{c}) \times \vec{a}=-\vec{b} \times \vec{a}$

$\Rightarrow(\vec{a} \times \vec{c}) \times \vec{a}=\vec{a} \times \vec{b}$

$\Rightarrow(\vec{a} \cdot \vec{a}) \vec{c}-(\vec{c} \cdot \vec{a}) \vec{a}=\vec{a} \times \vec{b}$

$\Rightarrow 2 \vec{c}-4 \vec{a}=\vec{a} \times \vec{b}$

Now $\quad \vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 0 \\ 1 & 1 & 1\end{array}\right|=-\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$

So, $2 \overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$

$=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$

$\Rightarrow \overrightarrow{\mathrm{c}}=\frac{3}{2} \hat{\mathrm{i}}-\frac{5}{2} \hat{\mathrm{j}}+\hat{\mathrm{k}}$

$|\vec{c}|=\sqrt{\frac{9}{4}+\frac{25}{4}+1}=\sqrt{\frac{38}{4}}=\sqrt{\frac{19}{2}}$

$|\overrightarrow{\mathrm{c}}|^{2}=\frac{19}{2}$