If the foot of the perpendicular drawn from ( | vedclass

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

Question

(A) The foot of the perpendicular from the origin $(0,0,0)$ to the plane is given as $P(1,2,3)$.Since the line segment joining the origin $(0,0,0)$ an

Vedclass · Accepted Answer

$x+2y+3z=14$

Vedclass · Answer

(A) The foot of the perpendicular from the origin $(0,0,0)$ to the plane is given as $P(1,2,3)$.Since the line segment joining the origin $(0,0,0)$ and the foot of the perpendicular $(1,2,3)$ is normal to the plane,the direction ratios of the normal to the plane are $(1-0, 2-0, 3-0) = (1, 2, 3)$.The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal direction ratios $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.Substituting the point $(1, 2, 3)$ and normal vector $(1, 2, 3)$ into the equation:$1(x-1) + 2(y-2) + 3(z-3) = 0$$x - 1 + 2y - 4 + 3z - 9 = 0$$x + 2y + 3z - 14 = 0$$x + 2y + 3z = 14$.

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

Similar Questions

The direction ratios of the normal to the plane passing through $(0,0,1)$,$(0,1,2)$,and $(1,0,3)$ are:

An equation of a plane parallel to the plane $x-2y+2z-5=0$ and which is at a distance of $1$ unit from the origin is:

The perpendicular distance of the origin from the plane $x-3y+4z-6=0$ is

$A$ plane meets the coordinate axes at points $A$, $B$ and $C$, such that the centroid of triangle $ABC$ is $(2, -\frac{2}{3}, \frac{1}{2})$. The perpendicular distance from the origin to this plane is:

The foot of the perpendicular drawn from the origin to the plane is $(4, -2, -5)$. Hence,the equation of the plane is

Vedclass Test Series

Exam Paper Generator

Online Exam Module