SATHEE: 3D Geometry 3 Question 42

3D Geometry 3 Question 42

####42. The point $P$ is the intersection of the straight line joining the points $Q (2, 3, 5)$ and $R (1, - 1, 4)$ with the plane $5 x - 4 y - z = 1$ . If $S$ is the foot of the perpendicular drawn from the point $T (2, 1, 4)$ to $Q R$ , then the length of the line segment $P S$ is

(2012)

(a) $\frac{1}{\sqrt{2}}$

(b) $\sqrt{2}$

(d) $2 \sqrt{2}$

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Answer:

Correct Answer: 42. (a)

Solution:

PLAN It is based on two concepts one is intersection of straight line and plane and other is the foot of perpendicular from a point to the straight line.

Description of Situation

(i) If the straight line

intersects the plane $A x + B y + C z + d = 0$ .

Then, $(a λ + x_{1}, b λ + y_{1}, c λ + z_{1})$ would satisfy

$A x + B y + C z + d = 0$

(ii) If $A$ is the foot of perpendicular from $P$ to $l$ .

Then, (DR’s of $P A$ ) is perpendicular to DR’s of $l$ .

$\Rightarrow \overset{―}{P A} \cdot \bar{l} = 0$

Equation of straight line $Q R$ , is

$\begin{aligned} \frac{x - 2}{1 - 2} = \frac{y - 3}{- 1 - 3} = \frac{z - 5}{4 - 5} \\ \Rightarrow \frac{x - 2}{- 1} = \frac{y - 3}{- 4} = \frac{z - 5}{- 1} \\ \Rightarrow \frac{x - 2}{1} = \frac{y - 3}{4} = \frac{z - 5}{1} = λ \\ ∴ P (λ + 2, 4 λ + 3, λ + 5) must lie on 5 x - 4 y - z = 1 . \\ \Rightarrow 5 (λ + 2) - 4 (4 λ + 3) - (λ + 5) = 1 \\ \Rightarrow 5 λ + 10 - 16 λ - 12 - λ - 5 = 1 \\ \Rightarrow - 7 - 12 λ = 1 \\ ∴ λ = \frac{- 2}{3} \\ or \\ P \frac{4}{3}, \frac{1}{3}, \frac{13}{3} \end{aligned}$

Again, we can assume $S$ from Eq.(i),

$as S (μ + 2, 4 μ + 3, μ + 5)$

$∴$ DR’s of $T S =< μ + 2 - 2, 4 μ + 3 - 1, μ + 5 - 4 >$

$=< μ, 4 μ + 2, μ + 1 >$

and DR’s of $Q R =< 1, 4, 1 >$

Since, perpendicular

$∴ 1 (μ) + 4 (4 μ + 2) + 1 (μ + 1) = 0$

$\Rightarrow μ = - \frac{1}{2}$ and $S \frac{3}{2}, 1, \frac{9}{2}$

$∴$ Length of $P S = \sqrt{\frac{3}{2} - {\frac{4}{3}}^{2} + 1 - {\frac{1}{3}}^{2} + \frac{9}{2} - {\frac{13}{3}}^{2}} = \frac{1}{\sqrt{2}}$

3D Geometry 3 Question 42

Answer:

Description of Situation

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3D Geometry 3 Question 42

Answer:

Description of Situation

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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