If (2, -1, 3) is the foot of the perpendicula | vedclass

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

Question

(A) Let the plane be $ax + by + cz + d = 0$. Since the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $(2, -1, 3)$,the normal v

Vedclass · Accepted Answer

$2x - y + 3z - 14 = 0$

Vedclass · Answer

(A) Let the plane be $ax + by + cz + d = 0$. Since the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $(2, -1, 3)$,the normal vector $\vec{n}$ to the plane is the vector from the origin to the foot of the perpendicular,which is $\vec{n} = (2 - 0)\hat{i} + (-1 - 0)\hat{j} + (3 - 0)\hat{k} = 2\hat{i} - \hat{j} + 3\hat{k}$. Thus,the equation of the plane is $2x - y + 3z = D$. Since the point $(2, -1, 3)$ lies on the plane,we substitute these coordinates into the equation: $2(2) - (-1) + 3(3) = D$ $4 + 1 + 9 = D$ $D = 14$. Therefore,the equation of the plane is $2x - y + 3z - 14 = 0$.

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:

Similar Questions

$A$ plane $\pi$ makes intercepts $3$ and $4$ respectively on $Z$-axis and $X$-axis. If $\pi$ is parallel to $Y$-axis, then its equation is:

If $P$ is the point $(2, 6, 3)$,then the equation of the plane passing through $P$ and perpendicular to $OP$,where $O$ is the origin,is:

$A$ 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 $(1, 2, 2)$.

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

If $\alpha$ is the acute angle between the planes $P_1$ and $P_2$,where the combined equation of the planes $P_1$ and $P_2$ is $2x^2 - 6y^2 - 12z^2 + 18yz + 2zx + xy = 0$,then the value of $\cos \alpha$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module