SATHEE: 3D Geometry 3 Question 66

3D Geometry 3 Question 66

####66. (i) Find the equation of the plane passing through the points $(2, 1, 0), (5, 0, 1)$ and $(4, 1, 1)$ .

(ii) If $P$ is the point $(2, 1, 6)$ , then the point $Q$ such that $P Q$ is perpendicular to the plane in (a) and the mid-point of $P Q$ lies on it.

$(2003, 4$ M)

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Answer:

(i) ( $x + y - 2_{z}$ )

(ii) $Q (6, 5, - 2)$

Solution:

(i) Equation of plane passing through $(2, 1, 0)$ is

$a (x - 2) + b (y - 1) + c (z - 0) = 0$

It also passes through $(5, 0, 1)$ and $(4, 1, 1)$ .

$\Rightarrow 3 a - b + c = 0$ and $2 a - 0 b + c = 0$

On solving, we get $\frac{a}{- 1} = \frac{b}{- 1} = \frac{c}{2}$

$∴$ Equation of plane is

$\begin{aligned} - (x - 2) - (y - 1) + 2 (z - 0) & = 0 \\ - (x - 2) - y + 1 + 2 z & = 0 \\ x \end{aligned}$

(ii) Let the coordinates of $Q$ be $(α, β, γ)$ .

Equation of line $P Q \Rightarrow \frac{x - 2}{1} = \frac{y - 1}{1} = \frac{z - 6}{- 2}$

Since, mid-point of $P$ and $Q$

$\frac{α + 2}{2}, \frac{β + 1}{2}, \frac{γ + 6}{2}$

which lies in line $P Q$ .

$\Rightarrow \frac{\frac{α - 2}{2} - 2}{2} = \frac{\frac{β + 1}{2} - 1}{1} = \frac{\frac{γ + 6}{2} - 6}{- 2}$

$\begin{matrix} = \frac{1 \frac{α + 2}{2} - 2 + 1 \frac{β + 1}{2} - 1 - 2 \frac{γ + 6}{2} - 6}{1 \cdot 1 + 1 \cdot 1 + (- 2) (- 2)} = 2 \\ ∵ \frac{α + 2}{2} - 1 \frac{β + 1}{2} - 2 \frac{γ + 6}{2} = 3 \\ \Rightarrow α = 6, β = 5, γ = - 2 \Rightarrow Q (6, 5, - 2) \end{matrix}$

3D Geometry 3 Question 66

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3D Geometry 3 Question 66

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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