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

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

Question

(B) The foot of the perpendicular is $P(2,3,-1)$ and the point from which the perpendicular is drawn is $A(4,2,1)$.Since the line segment $AP$ is perp

Vedclass · Accepted Answer

$2x - y + 2z + 1 = 0$

Vedclass · Answer

(B) The foot of the perpendicular is $P(2,3,-1)$ and the point from which the perpendicular is drawn is $A(4,2,1)$.Since the line segment $AP$ is perpendicular to the plane,the direction ratios of the normal to the plane are the same as the direction ratios of the line $AP$.The direction ratios of the normal are $(4-2, 2-3, 1-(-1)) = (2, -1, 2)$.Thus,the equation of the plane is of the form $2x - y + 2z + d = 0$.Since the plane passes through the point $(2,3,-1)$,we substitute these coordinates into the equation:$2(2) - (3) + 2(-1) + d = 0$$4 - 3 - 2 + d = 0$$-1 + d = 0$$d = 1$.Therefore,the equation of the plane is $2x - y + 2z + 1 = 0$.

If $(2,3,-1)$ is the foot of the perpendicular from $(4,2,1)$ to a plane,then the equation of the plane is

Similar Questions

If the planes $3x - 2y + 2z + 17 = 0$ and $4x + 3y - kz = 25$ are mutually perpendicular,then $k = $

Find the angle between the planes $ax + by + d = 0$ $(a^2 + b^2 \neq 0)$ and $z = 0$.

Let $\pi$ be the plane that passes through the point $(-2, 1, -1)$ and is parallel to the plane $2x - y + 2z = 0$. Then the foot of the perpendicular drawn from the point $(1, 2, 1)$ to the plane $\pi$ is

If the perpendicular distance from $(1, 2, 4)$ to the plane $2x + 2y - z + k = 0$ is $3$,then $k =$

If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are three non-coplanar vectors,then the vector equation $\overrightarrow{r}=(1-p-q) \overrightarrow{a}+p \overrightarrow{b}+q \overrightarrow{c}$ represents a :

Vedclass Test Series

Exam Paper Generator

Online Exam Module