The foot of the perpendicular drawn from the | vedclass

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

Question

(A) Let the origin be $O(0, 0, 0)$ and the foot of the perpendicular be $M(2, 1, -2)$.Since $OM$ is the normal to the plane,the normal vector $\vec{n}

Vedclass · Accepted Answer

$\bar{r} \cdot (2 \hat{i} + \hat{j} - 2 \hat{k}) = 9$

Vedclass · Answer

(A) Let the origin be $O(0, 0, 0)$ and the foot of the perpendicular be $M(2, 1, -2)$.Since $OM$ is the normal to the plane,the normal vector $\vec{n}$ is given by the vector $\vec{OM} = (2 - 0)\hat{i} + (1 - 0)\hat{j} + (-2 - 0)\hat{k} = 2\hat{i} + \hat{j} - 2\hat{k}$.The vector equation of a plane passing through a point with position vector $\vec{a}$ and having normal vector $\vec{n}$ is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.Here,$\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$.Thus,$\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = (2\hat{i} + \hat{j} - 2\hat{k}) \cdot (2\hat{i} + \hat{j} - 2\hat{k})$.Calculating the dot product: $(2 	imes 2) + (1 	imes 1) + (-2 	imes -2) = 4 + 1 + 4 = 9$.Therefore,the equation of the plane is $\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 9$.

The foot of the perpendicular drawn from the origin to a plane is $M(2, 1, -2)$. Find the vector equation of the plane.

Similar Questions

If the planes $\bar{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})=3$ and $\bar{r} \cdot(4 \hat{i}-\hat{j}+\mu \hat{k})=5$ are parallel,then the values of $\lambda$ and $\mu$ are respectively:

The Cartesian equation of the plane,passing through the points $(3,1,1)$,$(1,2,3)$ and $(-1,4,2)$,is

The equation of the plane passing through $(-1, 1, 2)$,whose normal makes equal acute angles with the coordinate axes is

The distance of a point $(1, 2, -1)$ from the plane $x - 2y + 4z + 10 = 0$ is

The equation of the plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module