The equations of planes parallel to the plane | vedclass

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

Question

(A) The equation of any plane parallel to $x+2y+2z+8=0$ is of the form $x+2y+2z+\lambda=0$.Given that the distance of this plane from the point $(1,1,

Vedclass · Accepted Answer

$x+2y+2z-13=0$ or $x+2y+2z-1=0$

Vedclass · Answer

(A) The equation of any plane parallel to $x+2y+2z+8=0$ is of the form $x+2y+2z+\lambda=0$.Given that the distance of this plane from the point $(1,1,2)$ is $2$ units.The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax+By+Cz+D=0$ is given by $d = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$.Substituting the values,we get:$2 = \frac{|1(1)+2(1)+2(2)+\lambda|}{\sqrt{1^2+2^2+2^2}}$$2 = \frac{|1+2+4+\lambda|}{\sqrt{1+4+4}}$$2 = \frac{|7+\lambda|}{\sqrt{9}}$$2 = \frac{|7+\lambda|}{3}$$|7+\lambda| = 6$This implies $7+\lambda = 6$ or $7+\lambda = -6$.Case $1$: $7+\lambda = 6 \Rightarrow \lambda = -1$.Case $2$: $7+\lambda = -6 \Rightarrow \lambda = -13$.Thus,the equations of the planes are $x+2y+2z-1=0$ and $x+2y+2z-13=0$.

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

Similar Questions

The direction cosines of the normal to the plane containing the lines having direction ratios $1, 2, 1$ and $4, 5, -3$ are

In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angles between them: $2x - 2y + 4z + 5 = 0$ and $3x - 3y + 6z - 1 = 0$.

The equation of the plane passing through the points $(1, 2, 3)$,$(-1, 4, 2)$ and $(3, 1, 1)$ is

If the foot of the perpendicular drawn from the point $P(2,0,-3)$ to the plane $\pi$ is $F(1,-2,0)$ and the equation of the plane $\pi$ is $ax+by-3z+d=0$, then $a+b+d=$

The coordinates of the points $A$ and $B$ are $(2, 3, 4)$ and $(-2, 5, -4)$ respectively. If a point $P(x, y, z)$ moves such that $PA^2 - PB^2 = k$,where $k$ is a constant,then the locus of $P$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module