Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

A vector parallel to the line of intersection of the planes $\vec{r}. (3\hat{i}-\hat{j}+\hat{k})=1 $ and $\vec{r}. (\hat{i}+4\hat{j}-2\hat{k})=2, $ is

Options:

$-2\hat{i}+7\hat{j}+13\hat{k}$

$2\hat{i}+7\hat{j}-13\hat{k}$

$-2\hat{i}-7\hat{j}+13\hat{k}$

$-2\hat{i}+7\hat{j}-13\hat{k}$

Correct Answer:

$-2\hat{i}+7\hat{j}+13\hat{k}$

Explanation:

The required vector is parallel to the vectors  $\vec{n_1}×\vec{n_2},$ where $\vec{n_1}$ and $\vec{n_2}$ are normal vectors to the given planes.

We have,

$\vec{n_1}=3\hat{i}-\hat{j}+\hat{k}$ and $\vec{n_2}= \hat{i}+4\hat{j}-2\hat{k}$

$∴\vec{n_1}×\vec{n_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\3 & -1 & 1\\1 & 4 & -2\end{vmatrix}= -2\hat{i}+7\hat{j}+13\hat{k}$

Hence, required vector is parallel to$ -2\hat{i}+7\hat{j}+13\hat{k}$