pi is a plane passing through the origin and | vedclass

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

Question

(B) The plane $\pi$ passes through the origin $(0, 0, 0)$ and contains two lines with direction ratios $\vec{n_1} = (1, -2, 2)$ and $\vec{n_2} = (2, 3

Vedclass · Accepted Answer

$2, 3, -1$

Vedclass · Answer

(B) The plane $\pi$ passes through the origin $(0, 0, 0)$ and contains two lines with direction ratios $\vec{n_1} = (1, -2, 2)$ and $\vec{n_2} = (2, 3, -1)$.The normal vector $\vec{n}$ to the plane $\pi$ is given by the cross product $\vec{n_1} 	imes \vec{n_2}$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 2 \ 2 & 3 & -1 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(-1 - 4) + \hat{k}(3 + 4) = -4\hat{i} + 5\hat{j} + 7\hat{k}$.Since the plane passes through the origin,its equation is $-4x + 5y + 7z = 0$.The line of intersection of the planes $x - y - z + 1 = 0$ (normal $\vec{n_3} = (1, -1, -1)$) and $\pi$ (normal $\vec{n} = (-4, 5, 7)$) has a direction vector $\vec{v} = \vec{n_3} 	imes \vec{n}$:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & -1 \ -4 & 5 & 7 \end{vmatrix} = \hat{i}(-7 + 5) - \hat{j}(7 - 4) + \hat{k}(5 - 4) = -2\hat{i} - 3\hat{j} + 1\hat{k}$.Thus,the direction ratios are proportional to $(-2, -3, 1)$,which is equivalent to $(2, 3, -1)$.

$\pi$ is a plane passing through the origin and containing two lines whose direction ratios are $1, -2, 2$ and $2, 3, -1$. Then,the direction ratios of the line of intersection of the planes $x - y - z + 1 = 0$ and $\pi$ are:

Similar Questions

Suppose the line $\frac{x-2}{\alpha}=\frac{y-2}{-5}=\frac{z+2}{2}$ lies on the plane $x+3y-2z+\beta=0$. Then $(\alpha+\beta)$ is equal to ... .

The angle between the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 3}{-2}$ and the plane $x + y + 4 = 0$ is ......... $^o$.

Find the equation of the line passing through the point $(3, 0, 1)$ and parallel to the planes $x+2y=0$ and $3y-z=0$.

The direction ratios (d.r.s.) of the normal to the plane passing through the origin and the line of intersection of the planes $x+2y+3z=4$ and $4x+3y+2z=1$ are

$A$ line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module