SATHEE: 3D Geometry 3 Question 9

3D Geometry 3 Question 9

####9. A plane passing through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $\frac{π}{4}$ with the plane $y - z + 5 = 0$ , also passes through the point

(2019 Main, 9 April I)

(a) $(\sqrt{2}, 1, 4)$

(b) $(- \sqrt{2}, 1, - 4)$

(d) $(\sqrt{2}, - 1, 4)$

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Answer:

(a)

Solution:

Let the equation of plane is

$a x + b y + c z = d$

Since plane (i) passes through the points $(0, - 1, 0)$ and $(0, 0, 1)$ , then $- b = d$ and $c = d$

$∴$ Equation of plane becomes $a x - d y + d z = d$ $∵$ The plane (ii) makes an angle of $\frac{π}{4}$ with the plane

$\begin{aligned} y - z + 5 & = 0 . \\ \cos \frac{π}{4} & = | \frac{- d - d}{\sqrt{a^{2} + d^{2} + d^{2}} \sqrt{1 + 1}} | \end{aligned}$

$[∵$ The angle between the two planes $a_{1} x + b_{1} y + c_{1} z + d = 0$ and $a_{2} x + b_{2} y + c_{2} z + d = 0$ is

$\cos θ = | \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}} |$

$\Rightarrow \frac{1}{\sqrt{2}} = \frac{| - 2 d |}{\sqrt{a^{2} + 2 d^{2}} \sqrt{2}} \Rightarrow \sqrt{a^{2} + 2 d^{2}} = | - d |$

$\Rightarrow a^{2} + 2 d^{2} = 4 d^{2}$

[squaring both sides]

$\Rightarrow a^{2} = 2 d^{2} \Rightarrow a = \pm \sqrt{2} d$

So, the Eq. (ii) becomes

$\pm \sqrt{2} x - y + z = 1$

Now, from options $(\sqrt{2}, 1, 4)$ satisfy the plane

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

3D Geometry 3 Question 9

Answer:

Solution:

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3D Geometry 3 Question 9

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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