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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

Which one of the following does not readily give oxygen upon heating?

Which one of the following has the smallest diameter?

The equation of the plane passing through the point $(1, 2, -3)$ and perpendicular to the planes $3x + y - 2z = 5$ and $2x - 5y - z = 7$ is:

Which one of the following statements is wrong?

Vedclass Test Series

Exam Paper Generator

Online Exam Module