Question

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

• A. $\overrightarrow{a}·\hat{i}+3=0$
• B. $\overrightarrow{a}·\hat{k}+4=0$
• C. $\overrightarrow{a}·\hat{i}+1=0$
• D. $\overrightarrow{a}·\hat{k}+2=0$

Answer 1

The correct option is D

$\overrightarrow{a}·\hat{k}+2=0$

Explanation for given data:

Step-1: Express the given data.

Given, $\overrightarrow{a}=\alpha \hat{i}+2\hat{j}+\beta \hat{k}\left(\alpha ,\beta \in R\right)...........\left(i\right)$,

$\overrightarrow{b}=\hat{i}+\hat{j}$ and $\overrightarrow{c}=\hat{i}-\hat{j}+4\hat{k}$. and $\overrightarrow{a}$ bisects the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$.

We know that If a vector bisect the angle between two vector then $\overrightarrow{a}=\lambda \left(\hat{b}+\hat{c}\right)\mathrm{or}\overrightarrow{\mathrm{a}}=\mu \left(\hat{\mathrm{b}}-\hat{\mathrm{c}}\right)$

Step-2: Consider $\overrightarrow{a}=\lambda \left(\hat{b}+\hat{c}\right)$

$\Rightarrow$$\overrightarrow{a}=\lambda \left(\frac{\left(\hat{i}+\hat{j}\right)}{\sqrt{2}}+\frac{\left(\hat{i}-\hat{j}+4\hat{k}\right)}{\sqrt{1^2+{\left(-1\right)}^2+4^2}}\right)$

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

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

$\Rightarrow$$\overrightarrow{a}=\frac{\lambda }{3\sqrt{2}}\left(4\hat{i}+2\hat{j}+4\hat{k}\right)............\left(ii\right)$

Compare equation $\left(i\right)\&\left(ii\right)$.

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

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

Put value of $\lambda$ in equation $\left(ii\right)$.

$\Rightarrow$$\overrightarrow{a}=\left(4\hat{i}+2\hat{j}+4\hat{k}\right)............\left(iii\right)$

Not follow the any option.

Step-3: Consider $\overrightarrow{a}=\mu \left(\hat{b}-\hat{c}\right)$

$\Rightarrow$$\overrightarrow{a}=\mu \left(\frac{\left(\hat{i}+\hat{j}\right)}{\sqrt{2}}-\frac{\left(\hat{i}-\hat{j}+4\hat{k}\right)}{\sqrt{1^2+{\left(-1\right)}^2+4^2}}\right)$

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

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

$\Rightarrow$$\overrightarrow{a}=\frac{\mu }{3\sqrt{2}}\left(2\hat{i}+4\hat{j}-4\hat{k}\right)............\left(iv\right)$

Compare equation $\left(i\right)\&\left(iv\right)$.

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

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

Put value of $\mu$ in equation $\left(iv\right)$.

$\Rightarrow$$\overrightarrow{a}=\left(\hat{i}+2\hat{j}-2\hat{k}\right)$

Taking Dot product with

$\Rightarrow$$\overrightarrow{a}·\hat{k}=\left(\hat{i}+2\hat{j}-2\hat{k}\right)·\hat{k}$

$\Rightarrow$$\overrightarrow{a}·\hat{k}=-2$

$\Rightarrow$$\overrightarrow{a}·\hat{k}+2=0$

Hence, option $\left(D\right)$ is correct.