The angle between the line of intersection of | vedclass

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

Question

(C) The direction vectors of the two planes are $\vec{n}_1 = 2\hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{n}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k}$.The dire

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(C) The direction vectors of the two planes are $\vec{n}_1 = 2\hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{n}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k}$.The direction vector $\vec{v}_1$ of the line of intersection of these two planes is given by the cross product $\vec{n}_1 	imes \vec{n}_2$:$\vec{v}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 2 & -3 \ 3 & 3 & -5 \end{vmatrix} = \hat{i}(-10 + 9) - \hat{j}(-10 + 9) + \hat{k}(6 - 6) = -\hat{i} + \hat{j} + 0\hat{k}$.The direction vector of the given line $r = 3\hat{i} + 2\hat{j} + \hat{k} + t(5\hat{i} + 5\hat{j} - 7\hat{k})$ is $\vec{v}_2 = 5\hat{i} + 5\hat{j} - 7\hat{k}$.The angle $	heta$ between the two lines is given by $\cos 	heta = \frac{|\vec{v}_1 \cdot \vec{v}_2|}{|\vec{v}_1| |\vec{v}_2|}$.Calculating the dot product: $\vec{v}_1 \cdot \vec{v}_2 = (-1)(5) + (1)(5) + (0)(-7) = -5 + 5 + 0 = 0$.Since the dot product is $0$,the angle $	heta$ is $\cos^{-1}(0) = \frac{\pi}{2}$.

The angle between the line of intersection of the two planes $r \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5$ and $r \cdot(3 \hat{i}+3 \hat{j}-5 \hat{k})=3$,and the line $r=3 \hat{i}+2 \hat{j}+\hat{k}+t(5 \hat{i}+5 \hat{j}-7 \hat{k})$ is

Similar Questions

If the image of the point $P(1, -2, 3)$ in the plane $2x + 3y - 4z + 22 = 0$ measured parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is $Q$,then $PQ$ is equal to:

The angle between the line $r = (2i - j + k) + \lambda (-i + j + k)$ and the plane $r \cdot (3i + 2j - k) = 4$ is:

The equation of the line of intersection of the planes $4x + 4y - 5z = 12$ and $8x + 12y - 13z = 32$ can be written as

The distance of the point whose position vector is $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $r \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=4$ is

The line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 1}{4}$ meets the plane $x + 2y + 3z = 14$ at the point:

Vedclass Test Series

Exam Paper Generator

Online Exam Module