If the distance of the point P(1, -2, 1) from | vedclass

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(A) The distance of point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz - D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C

Vedclass · Accepted Answer

$\left( \frac{8}{3}, \frac{4}{3}, -\frac{7}{3} \right)$

Vedclass · Answer

(A) The distance of point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz - D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}}$.Given point $P(1, -2, 1)$ and plane $x + 2y - 2z - \alpha = 0$,the distance is $5$.$5 = \frac{|1(1) + 2(-2) - 2(1) - \alpha|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{|1 - 4 - 2 - \alpha|}{\sqrt{9}} = \frac{|-5 - \alpha|}{3}$.Since $\alpha > 0$,$|-5 - \alpha| = |5 + \alpha| = 5 + \alpha$.$5 = \frac{5 + \alpha}{3} \implies 15 = 5 + \alpha \implies \alpha = 10$.The plane is $x + 2y - 2z - 10 = 0$.The line passing through $P(1, -2, 1)$ and perpendicular to the plane has direction ratios $(1, 2, -2)$.The equation of the line is $\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 1}{-2} = k$.Any point on the line is $(k + 1, 2k - 2, -2k + 1)$.This point lies on the plane: $(k + 1) + 2(2k - 2) - 2(-2k + 1) = 10$.$k + 1 + 4k - 4 + 4k - 2 = 10 \implies 9k - 5 = 10 \implies 9k = 15 \implies k = \frac{5}{3}$.Substituting $k$ back,the foot of the perpendicular is $(\frac{5}{3} + 1, 2(\frac{5}{3}) - 2, -2(\frac{5}{3}) + 1) = (\frac{8}{3}, \frac{4}{3}, -\frac{7}{3})$.

If the distance of the point $P(1, -2, 1)$ from the plane $x + 2y - 2z = \alpha$,where $\alpha > 0$,is $5$,then the foot of the perpendicular from $P$ to the plane is:

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