The coordinates of the foot of the perpendicu | vedclass

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

Question

(A) The equation of the plane is $2x + y - 2z = 18$. The normal vector to the plane is $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$. The length of the no

Vedclass · Accepted Answer

$(4, 2, -4)$

Vedclass · Answer

(A) The equation of the plane is $2x + y - 2z = 18$. The normal vector to the plane is $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$. The length of the normal vector is $|\vec{n}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$. Dividing the equation of the plane by $3$,we get the normal form $\frac{2}{3}x + \frac{1}{3}y - \frac{2}{3}z = 6$. The distance of the plane from the origin is $d = 6$. The coordinates of the foot of the perpendicular from the origin to the plane $ax + by + cz = d$ are given by $(\frac{ad}{a^2+b^2+c^2}, \frac{bd}{a^2+b^2+c^2}, \frac{cd}{a^2+b^2+c^2})$. Here,$a=2, b=1, c=-2$ and $d=18$. The denominator $a^2+b^2+c^2 = 9$. Thus,the coordinates are $(\frac{2 	imes 18}{9}, \frac{1 	imes 18}{9}, \frac{-2 	imes 18}{9}) = (4, 2, -4)$.

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x + y - 2z = 18$ are

Similar Questions

The point on the plane $2x - 2y + 4z + 5 = 0$ that is nearest to $\left(1, \frac{3}{2}, 2\right)$ is

The equation of the plane passing through $(2, 3, 4)$ and parallel to the plane $5x - 6y + 7z = 3$ is:

If the foot of the perpendicular drawn from the origin to a plane is $(1, 2, 3)$,then a point on that plane is

The planes: $2x - y + 4z = 5$ and $5x - 2.5y + 10z = 6$ are

The equation of the plane bisecting the line segment joining the points $(2,0,6)$ and $(-6,2,4)$ and perpendicular to it,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module