Discussion Forum

If OA= $\hat{i}+2\hat{j}+3\hat{k}$ and OB= $4 \hat{i}+\hat{k}$ are the position vectors of the points  A and B , then the position vector of a point  on the line passing through  B and parallel to the vector   OA x OB which is at a distance of $\sqrt{189}$  units from B is

Answer:

Explanation:

We have,

 OA=$\hat{i}+2\hat{j}+3\hat{k}$ and OB= $4 \hat{i}+\hat{k}$

 $\because$   $OA \times OB=\begin{bmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & 2&3 \\4 &0&1 \end{bmatrix}= 2 \hat{i}+11\hat{j}-8\hat{k}$

 equation of line passing through B and parallel to  OA x OB is

 r= $4\hat{i}+\hat{k}+\lambda  (2 \hat{i}+11 \hat{j}-8\hat{k})$

     r= $(4+2 \lambda)\hat{i}+11 \lambda \hat{j}+(1-8 \lambda) \hat{k}$

Distance from B to line  is $\sqrt {189}$

 $\therefore$    $\sqrt{189}$=  $\sqrt{(4+2\lambda-4)^{2}+(11\lambda)^{2}+(1-8 \lambda -1)^{2}}$

 189= $4\lambda^{2}+121\lambda^{2}+64 \lambda^{2}$

  189= 189 $\lambda ^{2}$

$\lambda$= $\pm$ 1

$\therefore$ Position vector ;ie on line  B

r= $6 \hat{i}+11\hat{j}-7 \hat{k}$