The equation of the plane passing through (-1 | vedclass

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

Question

(A) The equation of a plane passing through a point with position vector $\vec{a}$ and perpendicular to a normal vector $\vec{n}$ is given by $\vec{r}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of a plane passing through a point with position vector $\vec{a}$ and perpendicular to a normal vector $\vec{n}$ is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.Since the normal makes equal acute angles with the coordinate axes,its direction cosines are equal,i.e.,$l = m = n$. Since $l^2 + m^2 + n^2 = 1$,we have $3l^2 = 1$,so $l = m = n = \frac{1}{\sqrt{3}}$.Thus,the normal vector can be taken as $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.The given point is $\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$.Substituting these into the equation $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$:$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = (-\hat{i} + \hat{j} + 2\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k})$$= (-1)(1) + (1)(1) + (2)(1) = -1 + 1 + 2 = 2$.Therefore,the equation is $\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 2$.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

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

Similar Questions

In the following case,find the distance of the given point from the corresponding given plane.

Point Plane

$(-6, 0, 0)$ $2x - 3y + 6z - 2 = 0$

The equations of planes parallel to the plane $x+2y+2z+8=0$,which are at a distance of $2$ units from the point $(1,1,2)$ are

If $M$ denotes the midpoint of the line segment joining $A(4, 5, -10)$ and $B(-1, 2, 1)$, then the equation of the plane passing through $M$ and perpendicular to $AB$ is:

The equation of the plane passing through the points $(-1, -2, 0)$ and $(2, 3, 5)$ and parallel to the line $r = -3j + k + \lambda(2i + 5j - k)$ is

If the foot of the perpendicular drawn from the origin to a plane is $P(2,-1,4)$,then the equation of the plane is

Vedclass Test Series

Exam Paper Generator

Online Exam Module