A vector parallel to the line of intersection | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is perpendi

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is perpendicular to each of the normal vectors $\bar{n}_1 = 3 \hat{i}-\hat{j}+\hat{k}$ and $\bar{n}_2 = \hat{i}+4 \hat{j}-2 \hat{k}$.Therefore,the line is parallel to the vector $\bar{n}_1 	imes \bar{n}_2$.Calculating the cross product:$\bar{n}_1 	imes \bar{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 1 \ 1 & 4 & -2 \end{vmatrix}$$= \hat{i}((-1)(-2) - (1)(4)) - \hat{j}((3)(-2) - (1)(1)) + \hat{k}((3)(4) - (-1)(1))$$= \hat{i}(2 - 4) - \hat{j}(-6 - 1) + \hat{k}(12 + 1)$$= -2 \hat{i} + 7 \hat{j} + 13 \hat{k}$.

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

Similar Questions

If the points $(1, 1, \lambda)$ and $(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0$,then the integer value of $\lambda$ is:

$A$ plane $P$ contains the line $x+2y+3z+1=0=x-y-z-6$ and is perpendicular to the plane $-2x+y+z+8=0$. Then which of the following points lies on $P$?

The equation of a plane through the line of intersection of the planes $x + 2y = 3$ and $y - 2z + 1 = 0$,and perpendicular to the first plane $x + 2y = 3$ is:

The line passing through the point $2\bar{a}+\bar{b}$ and parallel to the vector $\bar{b}-\bar{c}$ and the plane passing through the point $\bar{a}$ and parallel to the vectors $\bar{b}+\bar{c}$ and $\bar{a}+2\bar{b}-\bar{c}$ intersect at $P$. The position vector of $P$ is

If $\theta$ is the angle between the line $\frac{x + 1}{3} = \frac{y - 2}{2} = \frac{z - 2}{4}$ and the plane $2x + y - 3z + 4 = 0$,then $64 \csc^2 \theta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module