SATHEE: 3D Geometry 3 Question 29

3D Geometry 3 Question 29

####29. If $L_{1}$ is the line of intersection of the planes $2 x - 2 y + 3 z - 2 = 0, x - y + z + 1 = 0$ and $L_{2}$ is the line of intersection of the planes $x + 2 y - z - 3 = 0$ , $3 x - y + 2 z - 1 = 0$ , then the distance of the origin from the plane, containing the lines $L_{1}$ and $L_{2}$ is (2018 Main)

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

(b) $\frac{1}{3 \sqrt{2}}$

(d) $\frac{1}{\sqrt{2}}$

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Answer:

Correct Answer: 29. (b)

Solution:

$L_{1}$ is the line of intersection of the plane $2 x - 2 y + 3 z - 2 = 0$ and $x - y + z + 1 = 0$ and $L_{2}$ is the line of intersection of the plane $x + 2 y - z - 3 = 0$ and $3 x - y + 2 z - 1 = 0$

Since $L_{i}$ is parallel to

$\hat{i} \hat{j} \hat{k}$

$2 - 2 3 = \hat{i} + \hat{j}$

$1 - 1 1$

Since $L_{2}$ is parallel to

$| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 1 \\ 3 & - 1 & 2 \end{matrix} |$ $= 3 \hat{i} - 5 \hat{j} - 7 \hat{k}$

Also, $L_{2}$ passes through $\frac{5}{7}, \frac{8}{7}, 0$ .

[put $z = 0$ in last two planes]

So, equation of plane is

$\begin{array}{ccc} x - \frac{5}{7} & y - \frac{8}{7} & z \\ 1 & 1 & 0 \\ 3 & - 5 & - 7 \end{array} = 0 \Rightarrow 7 x - 7 y + 8 z + 3 = 0$

Now, perpendicular distance from origin is

$\frac{3}{\sqrt{7^{2} + 7^{2} + 8^{2}}} = \frac{3}{\sqrt{162}} = \frac{1}{3 \sqrt{2}}$

3D Geometry 3 Question 29

Answer:

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3D Geometry 3 Question 29

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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