# Prove the following

Question:

A vector $\vec{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}(\alpha, \beta \in R)$ lies in the plane of the vectors, $\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}=\hat{i}-\hat{j}+4 \hat{k}$. If $\vec{a}$ bisects the angle between $\vec{b}$ and $\vec{c}$, then:

1. $\vec{a} \cdot \hat{i}+3=0$

2. $\vec{a} \cdot \hat{i}+1=0$

3. $\vec{a} \cdot \hat{k}+2=0$

4. $\vec{a} \cdot \hat{k}+4=0$

Correct Option: , 3

Solution:

Angle bisector between $\vec{b}$ and $\vec{c}$ can be

$\vec{a}=\lambda(\hat{b}+\hat{c}) \quad$ or $\quad \vec{a}=\mu(\hat{b}-\hat{c})$

If $\vec{a}=\lambda\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}+\frac{\hat{i}-\hat{j}+4 \hat{k}}{3 \sqrt{2}}\right)$

$=\frac{\lambda}{3 \sqrt{2}}[3 \hat{i}+3 \hat{j}+\hat{i}-\hat{j}+4 \hat{k}]$

$=\frac{\lambda}{3 \sqrt{2}}[4 \hat{i}+2 \hat{j}+4 \hat{k}]$

Compare with $\vec{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}$

$\frac{2 \lambda}{3 \sqrt{2}}=2 \quad \Rightarrow \quad \lambda=3 \sqrt{2}$

$\vec{a}=4 \hat{i}+2 \hat{j}+4 \hat{k}$

Not satisfy any option

Now consider $\vec{a}=\mu\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}-\frac{\hat{i}-\hat{j}+4 \hat{k}}{3 \sqrt{2}}\right)$

$\vec{a}=\frac{\mu}{3 \sqrt{2}}(3 i+3 \hat{j}-\hat{i}+\hat{j}-4 \hat{k})$

$=\frac{\mu}{3 \sqrt{2}}(2 \hat{i}+4 \hat{j}-4 \hat{k})$

Compare with $\vec{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}$

$\frac{2 \lambda}{3 \sqrt{2}}=2 \quad \Rightarrow \quad \lambda=3 \sqrt{2}$

$\vec{a}=4 \hat{i}+2 \hat{j}+4 \hat{k}$

Now consider $\vec{a}=\mu\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}-\frac{\hat{i}-\hat{j}+4 \hat{k}}{3 \sqrt{2}}\right)$

$\vec{a}=\frac{\mu}{3 \sqrt{2}}(3 \hat{i}+3 \hat{j}-\hat{i}+\hat{j}-4 \hat{k})$

$=\frac{\mu}{3 \sqrt{2}}(2 \hat{i}+4 \hat{j}-4 \hat{k})$

Compare with $\vec{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}$

$\frac{4 \mu}{3 \sqrt{2}}=2 \Rightarrow \mu=\frac{3 \sqrt{2}}{2}$

$\vec{a}=\hat{i}+2 \hat{j}-2 \hat{k}$

$\therefore \quad \vec{a} \cdot \vec{k}+2=0$

$-2+2=0$