SATHEE: 3D Geometry 3 Question 63

3D Geometry 3 Question 63

63. Find the equations of the plane containing the line $2 x - y + z - 3 = 0, 3 x + y + z = 5$ and at a distance of $\frac{1}{\sqrt{6}}$ from the point $(2, 1, - 1)$ .

$(2005, 2 M)$

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Answer:

Correct Answer: 63. $\frac{9}{2}$ cu units

Solution:

Equation of plane containing the lines

$\begin{aligned} 2 x - y + z - 3 = 0 and 3 x + y + z = 5 is \\ (2 x - y + z - 3) + λ (3 x + y + z - 5) = 0 \\ \Rightarrow (2 + 3 λ) x + (λ - 1) y + (λ + 1) z - 3 - 5 λ = 0 \end{aligned}$

Since, distance of plane from $(2, 1, - 1)$ to above plane is $1 / \sqrt{6}$ .

$∴ \frac{6 λ + 4 + λ - 1 - λ - 1 - 3 - 5 λ}{\sqrt{(3 λ + 2)^{2} + (λ - 1)^{2} + (λ + 1)^{2}}} = \frac{1}{\sqrt{6}}$

$\begin{aligned} \Rightarrow & 6 (λ - 1)^{2} & = 11 λ^{2} + 12 λ + 6 \\ \Rightarrow & λ & = 0, - \frac{24}{5} \end{aligned}$

$∴$ Equations of planes are

$2 x - y + z - 3 = 0 and 62 x + 29 y + 19 z - 105 = 0$

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3D Geometry 3 Question 63

63. Find the equations of the plane containing the line 2x−y+z−3=0,3x+y+z=5 and at a distance of 16 from the point (2,1,−1).

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63. Find the equations of the plane containing the line $2 x - y + z - 3 = 0, 3 x + y + z = 5$ and at a distance of $\frac{1}{\sqrt{6}}$ from the point $(2, 1, - 1)$ .