SATHEE: 3D Geometry 3 Question 8

3D Geometry 3 Question 8

####8. Let $P$ be the plane, which contains the line of intersection of the planes, $x + y + z - 6 = 0$ and $2 x + 3 y + z + 5 = 0$ and it is perpendicular to the $X Y$ -plane. Then, the distance of the point $(0, 0, 256)$ from $P$ is equal to

(2019 Main, 9 April II)

(a) $63 \sqrt{5}$

(b) $205 \sqrt{5}$

(d) $\frac{17}{\sqrt{5}}$

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Answer:

(c)

Solution:

Equation of plane, which contains the line of intersection of the planes

$\begin{matrix} x + y + z - 6 = 0 and 2 x + 3 y + z + 5 = 0, is \\ (x + y + z - 6) + λ (2 x + 3 y + z + 5) = 0 \end{matrix}$

$\Rightarrow (1 + 2 λ) x + (1 + 3 λ) y + (1 + λ) z + (5 λ - 6) = 0$

$∵$ The plane (i) is perpendicular to $X Y$ -plane (as DR’s of normal to $X Y$ -plane is $(0, 0, 1))$ .

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

$\Rightarrow λ = - 1$

On substituting $λ = - 1$ in Eq. (i),

we get

$- x - 2 y - 11 = 0$

$\Rightarrow x + 2 y + 11 = 0$

which is the required equation of the plane.

Now, the distance of the point $(0, 0, 256)$ from plane $P$ is

$\frac{0 + 0 + 11}{\sqrt{1 + 4}} = \frac{11}{\sqrt{5}}$

$[∵$ distance of $(x_{1}, y_{1}, z_{1})$ from the plane

$a x + b y + c z - d = 0, is | \frac{a x_{1} + b y_{1} + c z_{1} - d}{\sqrt{a^{2} + b^{2} + c^{2}}} |$

3D Geometry 3 Question 8

Answer:

Solution:

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3D Geometry 3 Question 8

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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