SATHEE: 3D Geometry 3 Question 3

3D Geometry 3 Question 3

####3. A plane which bisects the angle between the two given planes $2 x - y + 2 z - 4 = 0$ and $x + 2 y + 2 z - 2 = 0$ , passes through the point

(2019 Main, 12 April II)

(a) $(1, - 4, 1)$

(b) $(1, 4, - 1)$

(d) $(2, - 4, 1)$

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Answer:

(d)

Solution:

$\begin{aligned} Key Idea Equation of planes bisecting the angles between the & planes & a_{1} x + b_{1} y + c_{1} z + d_{1} = 0 and & a_{2} x + b_{2} y + c_{2} z + d_{2} = 0, are & \frac{a_{1} x + b_{1} y + c_{1} z + d_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}}} = \pm \frac{a_{2} x + b_{2} y + c_{2} z + d_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}} \end{aligned}$

Equation of given planes are

$and \begin{aligned} 2 x - y + 2 z - 4 & = 0 x + 2 y + 2 z - 2 & = 0 \end{aligned}$

Now, equation of planes bisecting the angles between the planes (i) and (ii) are

$\begin{aligned} \frac{2 x - y + 2 z - 4}{\sqrt{4 + 1 + 4}} = \pm \frac{x + 2 y + 2 z - 2}{\sqrt{1 + 4 + 4}} & \Rightarrow 2 x - y + 2 z - 4 = \pm (x + 2 y + 2 z - 2) \end{aligned}$

On taking (+ve) sign, we get a plane

$x - 3 y = 2$

On taking ( $- ve)$ sign, we get a plane

$3 x + y + 4 z = 6$

Now from the given options, the point $(2, - 4, 1)$ satisfy the plane of angle bisector $3 x + y + 4 z = 6$

3D Geometry 3 Question 3

Answer:

Solution:

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3D Geometry 3 Question 3

Answer:

Solution:

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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