Find the Cartesian equation of the plane vec{ | vedclass

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

Question

(C) The given equation of the plane is $\vec{r} = (1 + \lambda - \mu)\hat{i} + (2 - \lambda)\hat{j} + (3 - 2\lambda + 2\mu)\hat{k}$.We can rewrite thi

Vedclass · Accepted Answer

$2x + z = 5$

Vedclass · Answer

(C) The given equation of the plane is $\vec{r} = (1 + \lambda - \mu)\hat{i} + (2 - \lambda)\hat{j} + (3 - 2\lambda + 2\mu)\hat{k}$.We can rewrite this as $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} - 2\hat{k}) + \mu(-\hat{i} + 2\hat{k})$.This represents a plane passing through the point $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and parallel to the vectors $\vec{b} = \hat{i} - \hat{j} - 2\hat{k}$ and $\vec{c} = -\hat{i} + 2\hat{k}$.The normal vector $\vec{n}$ to the plane is given by $\vec{n} = \vec{b} 	imes \vec{c}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & -2 \ -1 & 0 & 2 \end{vmatrix} = \hat{i}(-2 - 0) - \hat{j}(2 - 2) + \hat{k}(0 - 1) = -2\hat{i} - \hat{k}$.The vector equation of the plane is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which implies $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.$\vec{a} \cdot \vec{n} = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (-2\hat{i} - \hat{k}) = (1)(-2) + (2)(0) + (3)(-1) = -2 - 3 = -5$.So,$\vec{r} \cdot (-2\hat{i} - \hat{k}) = -5$,or $\vec{r} \cdot (2\hat{i} + \hat{k}) = 5$.Substituting $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$,we get $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + \hat{k}) = 5$,which simplifies to $2x + z = 5$.

Find the Cartesian equation of the plane $\vec{r} = (1 + \lambda - \mu)\hat{i} + (2 - \lambda)\hat{j} + (3 - 2\lambda + 2\mu)\hat{k}$.

Similar Questions

$A$ plane passes through the point $(1, -2, 1)$ and is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. Find the distance of the plane from the point $(1, 2, 2)$.

Let the plane $\pi$ pass through the point $(1,0,1)$ and be perpendicular to the planes $2x+3y-z=2$ and $x-y+2z=1$. Let the equation of the plane passing through the point $(11,7,5)$ and parallel to the plane $\pi$ be $ax+by-z-d=0$. Then,$\frac{a}{b}+\frac{b}{d}=$

If $(2, -1, 3)$ is the foot of the perpendicular drawn from the origin $(0, 0, 0)$ to a plane,then the equation of that plane is:

The equation of the plane which bisects the line joining $(2, 3, 4)$ and $(6, 7, 8)$ is

If the plane $3x - 2y - z - 18 = 0$ meets the coordinate axes at $A, B, C$,then the centroid of $\triangle ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module