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Question

(A) Let the point be $P(7, 14, 5)$ and the plane be $2x + 4y - z - 2 = 0$.The length of the perpendicular $d$ from $(x_1, y_1, z_1)$ to $ax + by + cz

Vedclass · Accepted Answer

$3\sqrt{21}, (1, 2, 8)$

Vedclass · Answer

(A) Let the point be $P(7, 14, 5)$ and the plane be $2x + 4y - z - 2 = 0$.The length of the perpendicular $d$ from $(x_1, y_1, z_1)$ to $ax + by + cz + d = 0$ is given by $d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$.$d = \frac{|2(7) + 4(14) - 1(5) - 2|}{\sqrt{2^2 + 4^2 + (-1)^2}} = \frac{|14 + 56 - 5 - 2|}{\sqrt{4 + 16 + 1}} = \frac{63}{\sqrt{21}} = 3\sqrt{21}$.Let $M$ be the foot of the perpendicular. The line passing through $P$ perpendicular to the plane has direction ratios $(2, 4, -1)$.The equation of the line is $\frac{x - 7}{2} = \frac{y - 14}{4} = \frac{z - 5}{-1} = r$.So,$x = 2r + 7, y = 4r + 14, z = -r + 5$.Since $M$ lies on the plane $2x + 4y - z = 2$,we have $2(2r + 7) + 4(4r + 14) - (-r + 5) = 2$.$4r + 14 + 16r + 56 + r - 5 = 2 \implies 21r + 65 = 2 \implies 21r = -63 \implies r = -3$.Substituting $r = -3$,we get $x = 2(-3) + 7 = 1$,$y = 4(-3) + 14 = 2$,$z = -(-3) + 5 = 8$.Thus,the foot of the perpendicular is $(1, 2, 8)$.

Find the length of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$ and the coordinates of the foot of the perpendicular.

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