SATHEE: 3D Geometry 3 Question 50

3D Geometry 3 Question 50

####50. Consider a pyramid $O P Q R S$ located in the first octant $(x \geq 0, y \geq 0, z \geq 0)$ with $O$ as origin, and $O P$ and $O R$ along the $X$ -axis and the $Y$ -axis, respectively. The base $O P Q R$ of the pyramid is a square with $O P = 3$ . The point $S$ is directly above the mid-point $T$ of diagonal $O Q$ such that $T S = 3$ . Then,

(2016 Adv.)

(a) the acute angle between $O Q$ and $O S$ is $\frac{π}{3}$

(b) the equation of the plane containing the $△ O Q S$ is $x - y = 0$

(d) the perpendicular distance from $O$ to the straight line containing $R S$ is $\sqrt{\frac{15}{2}}$

Show Answer

Answer:

Correct Answer: 50. (b, c, d)

Solution:

Given, square base $O P = O R = 3$

$∴$

Also, mid-point of $O Q$ is $T \frac{3}{2}, \frac{3}{2}, 0$ .

Since, $S$ is directly above the mid-point $T$ of diagonal $O Q$ and $S T = 3$ . i.e. $S \frac{3}{2}, \frac{3}{2}, 3$

Here, DR’s of $O Q (3, 3, 0)$ and DR’s of $O S \frac{3}{2}, \frac{3}{2}, 3$ .

$∴ \cos θ = \frac{\frac{9}{2} + \frac{9}{2}}{\sqrt{9 + 9 + 0} \sqrt{\frac{9}{4} + \frac{9}{4} + 9}} = \frac{9}{\sqrt{18} \cdot \sqrt{\frac{27}{2}}} = \frac{1}{\sqrt{3}}$

$∴$ Option (a) is incorrect.

Now, equation of the plane containing the $△ O Q S$ is

$\begin{aligned} | \begin{array}{ccc} x & y & z \\ 3 & 3 & 0 \\ 3 / 2 & 3 / 2 & 3 \end{array} | = 0 \Rightarrow | \begin{array}{ccc} x & y & z \\ 1 & 1 & 0 \\ 1 & 1 & 2 \end{array} | = 0 \\ \Rightarrow x (2 - 0) - y (2 - 0) + z (1 - 1) = 0 \\ \Rightarrow 2 x - 2 y = 0 or x - y = 0 \\ ∴ Option (b) is correct. \end{aligned}$

Now, length of the perpendicular from $P (3, 0, 0)$ to the plane containing $△ O Q S$ is $\frac{| 3 - 0 |}{\sqrt{1 + 1}} = \frac{3}{\sqrt{2}}$

$∴$ Option (c) is correct.

Here, equation of $R S$ is

$\begin{aligned} \Rightarrow & \frac{x - 0}{3 / 2} & = \frac{y - 3}{- 3 / 2} = \frac{z - 0}{3} = λ \\ x & = \frac{3}{2} λ, y = - \frac{3}{2} λ + 3, z = 3 λ \end{aligned}$

To find the distance from $O (0, 0, 0)$ to $R S$ .

Let $M$ be the foot of perpendicular.

$∵ \overset{―}{O M} ⊥ \overset{―}{R S} \Rightarrow \overset{―}{O M} \cdot \overset{―}{R S} = 0$

$\Rightarrow \frac{9 λ}{4} - \frac{3}{2} 3 - \frac{3 λ}{2} + 3 (3 λ) = 0 \Rightarrow λ = \frac{1}{3}$

$∴ M \frac{1}{2}, \frac{5}{2}, 1 \Rightarrow O M = \sqrt{\frac{1}{4} + \frac{25}{4} + 1} = \sqrt{\frac{30}{4}} = \sqrt{\frac{15}{2}}$

$∴$ Option (d) is correct.

3D Geometry 3 Question 50

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

3D Geometry 3 Question 50

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu