The equation of the plane passing through (-2 | vedclass

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

Question

(B) The equation of a plane passing through $(-2, 2, 2)$ is given by $a(x + 2) + b(y - 2) + c(z - 2) = 0$. Since the plane passes through $(2, -2, -2)

Vedclass · Accepted Answer

$5x + 3y + 2z = 0$

Vedclass · Answer

(B) The equation of a plane passing through $(-2, 2, 2)$ is given by $a(x + 2) + b(y - 2) + c(z - 2) = 0$. Since the plane passes through $(2, -2, -2)$,we substitute these coordinates into the equation: $a(2 + 2) + b(-2 - 2) + c(-2 - 2) = 0 \Rightarrow 4a - 4b - 4c = 0 \Rightarrow a - b - c = 0$ (Equation $1$). The plane is perpendicular to $9x - 13y - 3z = 0$,so the normal vector $(a, b, c)$ is perpendicular to the normal vector $(9, -13, -3)$ of the given plane. Thus,$9a - 13b - 3c = 0$ (Equation $2$). Solving equations $(1)$ and $(2)$ using cross-multiplication: $\frac{a}{(-1)(-3) - (-1)(-13)} = \frac{-b}{(1)(-3) - (-1)(9)} = \frac{c}{(1)(-13) - (-1)(9)}$ $\frac{a}{3 - 13} = \frac{-b}{-3 + 9} = \frac{c}{-13 + 9}$ $\frac{a}{-10} = \frac{-b}{6} = \frac{c}{-4} \Rightarrow \frac{a}{5} = \frac{b}{3} = \frac{c}{2}$. Substituting the direction ratios $(5, 3, 2)$ into the plane equation: $5(x + 2) + 3(y - 2) + 2(z - 2) = 0$ $5x + 10 + 3y - 6 + 2z - 4 = 0$ $5x + 3y + 2z = 0$.

The equation of the plane passing through $(-2, 2, 2)$ and $(2, -2, -2)$ and perpendicular to the plane $9x - 13y - 3z = 0$ is

Similar Questions

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

Equation of planes parallel to the plane $x-2y+2z+4=0$ which are at a distance of one unit from the point $(1,2,3)$ are

The sum of the intercepts on the coordinate axes of the plane passing through the point $(-2, -2, 2)$ and containing the line joining the points $(1, -1, 2)$ and $(1, 1, 1)$ is

If $A$ and $B$ are the feet of the perpendiculars drawn from the point $Q(a, b, c)$ to the planes $YZ$ and $ZX$ respectively,then the equation of the plane passing through the points $A, B$,and $O$ is (where $O$ is the origin).

Find the equation of the plane passing through the points $(2, 2, 1)$ and $(9, 3, 6)$ and perpendicular to the plane $2x + 6y + 6z - 1 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module