Question

A unit vector in the plane of $i+2j+k$ and $i+j+2k$ and perpendicular to $2i+j+k$ is?

• A. $j-k$
• B. $\frac{\left(i+j\right)}{\sqrt{2}}$
• C. $\frac{\left(j-k\right)}{\sqrt{2}}$
• D. $5\left(j-k\right)$

Answer 1

The correct option is C

$\frac{\left(j-k\right)}{\sqrt{2}}$

Explanation for correct option:

Step 1. Finding the equations by using the given data:

Given,

$\overrightarrow{a}=i+2j+k$

$\overrightarrow{b}=i+j+2k$

$\overrightarrow{c}=2i+j+k$

Let the required unit vector $\overrightarrow{m}=ai+bj+ck$

Since, $\overrightarrow{m}$ is a unit vector, therefore $a^2+b^2+c^2=1.........\left(i\right)$

$\overrightarrow{m}$ is perpendicular to vector $2i+j+k$

$\Rightarrow$$2a+b+c=0.........\left(ii\right)$

$\overrightarrow{m}$ is perpendicular to the normal to the plane.

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

$\Rightarrow$ $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}i& j& k\\ 1& 2& 1\\ 1& 1& 2\end{array}\right|$

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

$\Rightarrow$ $\overrightarrow{a}\times \overrightarrow{b}=3i-j-k$

$\Rightarrow$ $3a-b-c=0.......\left(iii\right)$

Step 2. Solving equation equation $\left(i\right),\left(ii\right)\&\left(iii\right)$.

From equation $\left(ii\right)$

$\Rightarrow$$b+c=-2a$

Put value of $b+c$ in equation $\left(iii\right)$

$\Rightarrow$ $a=0$

and $b=-c$

$\Rightarrow$ $b=\frac{1}{\sqrt{2}}$

and $c=-\frac{1}{\sqrt{2}}$

Therefore, required vector is $\overrightarrow{m}=\frac{1}{\sqrt{2}}j-\frac{1}{\sqrt{2}}k$

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