SATHEE: 3D Geometry 3 Question 43

3D Geometry 3 Question 43

####43. If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is 5 , then the foot of the perpendicular form $P$ to the plane is

(2010)

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

(b) $\frac{4}{3}, \frac{4}{3}, \frac{1}{3}$

(d) $\frac{2}{3}, \frac{1}{3}, \frac{5}{2}$

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Answer:

Correct Answer: 43. (a)

Solution:

Distance of point $P$ from plane $= 5$

$\begin{array}{ll} ∴ & 5 = | \frac{1 - 4 - 2 - α}{3} | \\ \Rightarrow & α = 10 \end{array}$

Foot of perpendicular

$\begin{aligned} \frac{x - 1}{1} & = \frac{y + 2}{2} = \frac{z - 1}{- 2} = \frac{5}{3} \\ \Rightarrow x & = \frac{8}{3}, y = \frac{4}{3}, z = - \frac{7}{3} \end{aligned}$

Thus, the foot of the perpendicular is $A \frac{8}{3}, \frac{4}{3}, - \frac{7}{3}$ .

3D Geometry 3 Question 43

Answer:

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3D Geometry 3 Question 43

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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