SATHEE: 3D Geometry 3 Question 67

3D Geometry 3 Question 67

####67. If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $\sqrt{6}$ , then $| d |$ is equal to….

(2010)

Show Answer

Answer:

$| D | = 6$

Solution:

Equation of the plane containing the lines $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$

$is a (x - 2) + b (y - 3) + c (z - 4) = 0$

where, $3 a + 4 b + 5 c = 0$

$2 a + 3 b + 4 c = 0$

and $a (1 - 2) + b (2 - 3) + c (2 - 3) = 0$

i.e.

$a + b + c = 0$

From Eqs. (ii) and (iii), $\frac{a}{1} = \frac{b}{- 2} = \frac{c}{1}$ , which satisfy

Eq. (iv).

Plane through lines is $x - 2 y + z = 0$ .

Given plane is $A x - 2 y + z = d$ is $\sqrt{6}$ .

$∴$ Planes must be parallel, so $A = 1$ and then

$\frac{| d |}{\sqrt{6}} = \sqrt{6} \Rightarrow | d | = 6$

3D Geometry 3 Question 67

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 67

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu