Find the equation of the plane passing throug | vedclass

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

Question

(B) The equation of the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\vec{r} \cdot (2\

Vedclass · Accepted Answer

$\vec{r} \cdot (3\hat{i} + 2\hat{j} + 0\hat{k}) = 8$

Vedclass · Answer

(B) The equation of the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$ is given by:$[\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) - 5] + \lambda [\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) - 3] = 0$$\Rightarrow \vec{r} \cdot [(1 + 2\lambda)\hat{i} + (3 - \lambda)\hat{j} + (-1 + \lambda)\hat{k}] = 5 + 3\lambda \quad (i)$Since the plane passes through the point $(2, 1, -2)$,the position vector is $\vec{r} = 2\hat{i} + \hat{j} - 2\hat{k}$. Substituting this into equation $(i)$:$(2\hat{i} + \hat{j} - 2\hat{k}) \cdot [(1 + 2\lambda)\hat{i} + (3 - \lambda)\hat{j} + (-1 + \lambda)\hat{k}] = 5 + 3\lambda$$2(1 + 2\lambda) + 1(3 - \lambda) - 2(-1 + \lambda) = 5 + 3\lambda$$2 + 4\lambda + 3 - \lambda + 2 - 2\lambda = 5 + 3\lambda$$7 + \lambda = 5 + 3\lambda$$2 = 2\lambda \Rightarrow \lambda = 1$Substituting $\lambda = 1$ in equation $(i)$:$\vec{r} \cdot [(1 + 2(1))\hat{i} + (3 - 1)\hat{j} + (-1 + 1)\hat{k}] = 5 + 3(1)$$\vec{r} \cdot (3\hat{i} + 2\hat{j} + 0\hat{k}) = 8$.

Find the equation of the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$,and passing through the point $(2, 1, -2)$.

Similar Questions

The equation of the plane passing through the points $(2, 3, 1)$ and $(4, -5, 3)$ and parallel to the $X$-axis is:

The perpendicular distance from the origin to the plane $x + 2y - 2z + 5 = 0$ equals $.........$ units.

Find the equation of the plane passing through the point $(-1, 3, 2)$ and perpendicular to each of the planes $x + 2y + 3z = 5$ and $3x + 3y + z = 0$.

$A$ tetrahedron has vertices $P(1,2,1), Q(2,1,3), R(-1,1,2)$ and $O(0,0,0)$. Then the angle between the faces $OPQ$ and $PQR$ is

If the plane $\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1$ cuts the coordinate axes at points $A, B$,and $C$,then the area of the triangle $ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module