SATHEE: 3D Geometry 3 Question 4

3D Geometry 3 Question 4

####4. If the line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{- 1}$ intersects the plane $2 x + 3 y - z + 13 = 0$ at a point $P$ and the plane $3 x + y + 4 z = 16$ at a point $Q$ , then $P Q$ is equal to

(2019 Main, 12 April I)

(a) 14

(b) $\sqrt{14}$

(d) $2 \sqrt{14}$

Show Answer

Answer:

(d)

Solution:

Equation of given line is

$\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{- 1} = r (let)$

Now, coordinates of a general point over given

line is $R (3 r + 2, 2 r - 1, - r + 1)$

Let the coordinates of point $P$ are $(3 r_{1} + 2, 2 r_{1} - 1, r_{1} + 1)$ and $Q$ are $(3 r_{2} + 2, 2 r_{2} - 1, - r_{2} + 1)$ .

Since, $P$ is the point of intersection of line (i) and the plane $2 x + 3 y - z + 13 = 0$ , so

$\begin{aligned} 2 (3 r_{1} + 2) + 3 (2 r_{1} - 1) - (- r_{1} + 1) + 13 = 0 \\ \Rightarrow & 6 r_{1} + 4 + 6 r_{1} - 3 + r_{1} - 1 + 13 = 0 \\ \Rightarrow & 13 r_{1} + 13 = 0 \Rightarrow r_{1} = - 1 \end{aligned}$

So, point $P (- 1, - 3, 2)$

And, similarly for point ’ $Q$ ‘, we get

$\begin{aligned} 3 (3 r_{2} + 2) + (2 r_{2} - 1) + 4 (- r_{2} + 1) = 16 \\ \Rightarrow & 7 r_{2} = 7 \Rightarrow r_{2} = 1 \end{aligned}$

So, point is $Q (5, 1, 0)$

$Now, \begin{aligned} P Q & = \sqrt{(5 + 1)^{2} + (1 + 3)^{2} + 2^{2}} \\ = \sqrt{36 + 16 + 4} \\ = \sqrt{56} = 2 \sqrt{14} \end{aligned}$

3D Geometry 3 Question 4

Answer:

Solution:

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 4

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu