SATHEE: 3D Geometry 3 Question 41

3D Geometry 3 Question 41

####41. The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $2 / \sqrt{3}$ from the point $(3, 1, - 1)$ is (2012)

(a) $5 x - 11 y + z = 17$

(b) $\sqrt{2} x + y = 3 \sqrt{2} - 1$

(d) $x - \sqrt{2} y = 1 - \sqrt{2}$

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Answer:

Correct Answer: 41. (a)

Solution:

Key Idea

(i) Equation of plane through intersection of two planes,

$\begin{aligned} i.e. (a_{1} x + b_{1} y + c_{1} z + d_{1}) + λ & (a_{2} x + b_{2} y + c_{2} z + d_{2}) = 0 \end{aligned}$

(ii) Distance of a point $(x_{1}, y_{1}, z_{1})$ from

$\begin{aligned} a x + b y + c z + d & = 0 & = \frac{| a x_{1} + b y_{1} + c z_{1} + d |}{\sqrt{a^{2} + b^{2} + c^{2}}} \end{aligned}$

Equation of plane passing through intersection of two planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ is

$(x + 2 y + 3 z - 2) + λ (x - y + z - 3) = 0$

$\Rightarrow (1 + λ) x + (2 - λ) y + (3 + λ) z - (2 + 3 λ) = 0$ whose distance from $(3, 1, - 1)$ is $\frac{2}{\sqrt{3}}$ .

$\Rightarrow \frac{| 3 (1 + λ) + 1 \cdot (2 - λ) - 1 (3 + λ) - (2 + 3 λ) |}{\sqrt{(1 + λ)^{2} + (2 - λ)^{2} + (3 + λ)^{2}}} = \frac{2}{\sqrt{3}}$

$\Rightarrow \frac{| - 2 λ |}{\sqrt{3 λ^{2} + 4 λ + 14}} = \frac{2}{\sqrt{3}}$

$\Rightarrow 3 λ^{2} = 3 λ^{2} + 4 λ + 14$

$\Rightarrow λ = - \frac{7}{2}$

$∴ 1 - \frac{7}{2} x + 2 + \frac{7}{2} y + 3 - \frac{7}{2} z - 2 - \frac{21}{2} = 0$

$\Rightarrow - \frac{5 x}{2} + \frac{11}{2} y - \frac{1}{2} z + \frac{17}{2} = 0$

or $5 x - 11 y + z - 17 = 0$

3D Geometry 3 Question 41

Answer:

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3D Geometry 3 Question 41

Answer:

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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