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

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(A) Let the equation of the plane be $A(x + 2) + B(y - 2) + C(z - 2) = 0$. Since it passes through $(2, -2, -2)$,we have $A(2 + 2) + B(-2 - 2) + C(-2

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the equation of the plane be $A(x + 2) + B(y - 2) + C(z - 2) = 0$. Since it passes through $(2, -2, -2)$,we have $A(2 + 2) + B(-2 - 2) + C(-2 - 2) = 0$,which simplifies to $4A - 4B - 4C = 0$,or $A - B - C = 0$ ... $(i)$. The plane is perpendicular to $9x - 13y - 3z = 0$,so the normal vector of the required plane $(A, B, C)$ is perpendicular to the normal vector of the given plane $(9, -13, -3)$. Thus,$9A - 13B - 3C = 0$ ... $(ii)$. Solving $(i)$ and $(ii)$ using cross product: $\frac{A}{(-1)(-3) - (-1)(-13)} = \frac{B}{(-1)(9) - (1)(-3)} = \frac{C}{(1)(-13) - (-1)(9)}$ $\frac{A}{3 - 13} = \frac{B}{-9 + 3} = \frac{C}{-13 + 9}$ $\frac{A}{-10} = \frac{B}{-6} = \frac{C}{-4}$ Simplifying the ratios,we get $A:B:C = 5:3:2$. Substituting these 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

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