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(B) Let $Q = (1, 2, 3)$ be the point and $R = (-1, 3, -2)$ be the foot of the perpendicular on the plane.The normal vector $\vec{n}$ to the plane is t

Vedclass · Accepted Answer

$\sqrt{\frac{15}{2}}$

Vedclass · Answer

(B) Let $Q = (1, 2, 3)$ be the point and $R = (-1, 3, -2)$ be the foot of the perpendicular on the plane.The normal vector $\vec{n}$ to the plane is the vector $\vec{QR}$.$\vec{n} = \vec{R} - \vec{Q} = (-1 - 1, 3 - 2, -2 - 3) = (-2, 1, -5)$.We can also take the normal vector as $\vec{n} = (2, -1, 5)$.The equation of the plane passing through $R(-1, 3, -2)$ with normal vector $\vec{n} = (2, -1, 5)$ is given by:$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$$2(x - (-1)) - 1(y - 3) + 5(z - (-2)) = 0$$2(x + 1) - (y - 3) + 5(z + 2) = 0$$2x + 2 - y + 3 + 5z + 10 = 0$$2x - y + 5z + 15 = 0$.The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.Here, $A = 2, B = -1, C = 5, D = 15$.$d = \frac{|15|}{\sqrt{2^2 + (-1)^2 + 5^2}} = \frac{15}{\sqrt{4 + 1 + 25}} = \frac{15}{\sqrt{30}}$.Rationalizing the denominator:$d = \frac{15}{\sqrt{30}} 	imes \frac{\sqrt{30}}{\sqrt{30}} = \frac{15\sqrt{30}}{30} = \frac{\sqrt{30}}{2} = \sqrt{\frac{30}{4}} = \sqrt{\frac{15}{2}}$.

Let the foot of the perpendicular drawn from the point $(1, 2, 3)$ to a plane be $(-1, 3, -2)$. Then the perpendicular distance from the origin to the plane is

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