Question

Given that $i+j,i+2j$ are two vectors. Let a unit vector coplanar with these vectors and perpendicular to the plane of the two given vectors be $\frac{1}{\sqrt{k}}(i+mj)$ .Find $k+m$?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C 3

$\hat{i}+\hat{j},\hat{i}+2\hat{j}and\hat{i}+m\hat{j}$

are three coplanar vector

$\overrightarrow{\hat{i}+\hat{j}}\cdot \left(\right(\hat{i}+2\hat{j})\times (\hat{i}+m\hat{j}\left)\right)=0$

$\left|\begin{array}{ccc}1& 1& 0\\ 1& 2& 0\\ 1& m& 0\end{array}\right|=0$

$1\left(0\right)-1\left(0\right)+0=0$

$0=0$

hence it is coplanar

now vector $\hat{i}+m\hat{j}$ is perpendicular to planes contain two given point

$(\hat{i}\times \hat{j})\times (\hat{i}+m\hat{j})=0$

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 1& m& 0\end{array}\right|=0$

$\hat{i}\left(0\right)-\hat{j}\left(0\right)+\hat{k}(m-1)=0$

$(m-1)\hat{k}=0$

comparing both sides

$m-1=0$

$m=1$

$\frac{1}{\sqrt{k}}(\hat{i}+m\hat{j})=\frac{1}{\sqrt{1^2+1^2}}(\hat{i}+\hat{j})$ here m=1

$\frac{1}{\sqrt{k}}(\hat{i}+m\hat{j})=\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$

comparing both sides

$k=2,m=1$

$k+m=2+1=3$