If a plane pi passes through the point (-1,6, | vedclass

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(D) The equation of a plane passing through the point $(x_1, y_1, z_1)$ and perpendicular to two planes with normal vectors $\vec{n_1}$ and $\vec{n_2}

Vedclass · Accepted Answer

$\sqrt{29}$

Vedclass · Answer

(D) The equation of a plane passing through the point $(x_1, y_1, z_1)$ and perpendicular to two planes with normal vectors $\vec{n_1}$ and $\vec{n_2}$ is given by the determinant form:$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = 0$Substituting the given values:$\begin{vmatrix} x+1 & y-6 & z-2 \ 1 & 2 & 2 \ 3 & 3 & 2 \end{vmatrix} = 0$Expanding the determinant:$(x+1)(4-6) - (y-6)(2-6) + (z-2)(3-6) = 0$$-2(x+1) + 4(y-6) - 3(z-2) = 0$$-2x - 2 + 4y - 24 - 3z + 6 = 0$$-2x + 4y - 3z - 20 = 0$ or $2x - 4y + 3z + 20 = 0$The perpendicular distance $d$ from a point $(x_0, y_0, z_0)$ to a plane $Ax+By+Cz+D=0$ is given by $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.For the point $(1, -1, 1)$ and the plane $2x - 4y + 3z + 20 = 0$:$d = \frac{|2(1) - 4(-1) + 3(1) + 20|}{\sqrt{2^2 + (-4)^2 + 3^2}}$$d = \frac{|2 + 4 + 3 + 20|}{\sqrt{4 + 16 + 9}} = \frac{29}{\sqrt{29}} = \sqrt{29}$

If a plane $\pi$ passes through the point $(-1,6,2)$ and is perpendicular to the planes $x+2y+2z-5=0$ and $3x+3y+2z-8=0$, then the perpendicular distance from the point $(1,-1,1)$ to the plane $\pi$ is

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