SATHEE: 3D Geometry 3 Question 11

3D Geometry 3 Question 11

####11. The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ , which is perpendicular to the plane $x - y + z = 0$ is

(2019 Main, 8 April II)

(a) $r \cdot (\hat{i} - \hat{k}) - 2 = 0$

(b) $r \times (\hat{i} + \hat{k}) + 2 = 0$

(d) $r \cdot (\hat{i} - \hat{k}) + 2 = 0$

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Answer:

(d)

Solution:

Since, equation of planes passes through the line of intersection of the planes

$x + y + z = 1$

and $2 x + 3 y + 4 z = 5$ , is

$(x + y + z - 1) + λ (2 x + 3 y + 4 z - 5) = 0$

$\Rightarrow (1 + 2 λ) x + (1 + 3 λ) y + (1 + 4 λ) z - (1 + 5 λ) = 0$

$∵$ The plane (i) is perpendicular to the plane

$x - y + z = 0 .$

$∴ (1 + 2 λ) - (1 + 3 λ) + (1 + 4 λ) = 0$

$[∵$ if plane $a_{1} x + b_{1} y + c_{1} z + d_{1} = 0$ is perpendicular to plane $a_{2} x + b_{2} y + c_{2} z + d_{2} = 0$ , then $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0]$

$\Rightarrow 3 λ + 1 = 0$

$\Rightarrow λ = - \frac{1}{3}$

So, the equation of required plane, is

$\begin{aligned} 1 - \frac{2}{3} x + 1 - \frac{3}{3} y + 1 - \frac{4}{3} z - 1 - \frac{5}{3} = 0 \\ \Rightarrow \frac{1}{3} x - \frac{1}{3} z + \frac{2}{3} = 0 \Rightarrow x - z + 2 = 0 \end{aligned}$

Now, vector form, is $r \cdot (\hat{i} - \hat{k}) + 2 = 0$

3D Geometry 3 Question 11

Answer:

Solution:

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3D Geometry 3 Question 11

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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