SATHEE: 3D Geometry 3 Question 30

3D Geometry 3 Question 30

33. The distance of the point $(1, 3, - 7)$ from the plane passing through the point $(1, - 1, - 1)$ having normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$ , is

(a) $\frac{20}{\sqrt{74}}$ units

(b) $\frac{10}{\sqrt{83}}$ units

(d) $\frac{10}{\sqrt{74}}$ units

(2017 Main)

Show Answer

Answer:

Correct Answer: 33. (b)

Solution:

Given, equations of lines are

$\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$

and

$\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$

Let $n_{1} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $n_{2} = 2 \hat{i} - \hat{j} - \hat{k}$ $∴$ Any vector $n$ perpendicular to both $n_{1}, n_{2}$ is given by

$n = n_{1} \times n_{2}$

$\Rightarrow n = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} 1 & - 2 & 3 2 & - 1 & - 1 \end{array} | = 5 \hat{i} + 7 \hat{j} + 3 \hat{k}$

$∴$ Equation of a plane passing through $(1, - 1, - 1)$ and perpendicular to $n$ is given by

$\begin{aligned} 5 (x - 1) + 7 (y + 1) + 3 (z + 1) = 0 \\ \Rightarrow 5 x + 7 y + 3 z + 5 = 0 \\ ∴ Required distance = | \frac{5 + 21 - 21 + 5}{\sqrt{5^{2} + 7^{2} + 3^{2}}} | = \frac{10}{\sqrt{83}} units \end{aligned}$

3D Geometry 3 Question 30

33. The distance of the point $(1, 3, - 7)$ from the plane passing through the point $(1, - 1, - 1)$ having normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$ , is

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 30

33. The distance of the point (1,3,−7) from the plane passing through the point (1,−1,−1) having normal perpendicular to both the lines x−11=y+2−2=z−43 and x−22=y+1−1=z+7−1, is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

33. The distance of the point $(1, 3, - 7)$ from the plane passing through the point $(1, - 1, - 1)$ having normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$ , is