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Three Dimensional Geometry

Long Answer Type

201. Find the vector equation of the plane passing through the intersection of the planes:

straight r with rightwards arrow on top. space open parentheses straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top close parentheses space equals space 6 comma space space space space straight r with rightwards arrow on top. space space open parentheses 2 space straight i with hat on top space plus space space 3 space straight j with hat on top space plus space 4 space straight k with hat on top close parentheses space equals space minus 5

and the point (1, 1, 1).

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202. Find the vector equation of the plane passing through the intersection of the planes

straight r with rightwards arrow on top. space open parentheses 2 straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses space equals 7 comma space space space space space straight r with rightwards arrow on top space. space open parentheses 2 space straight i with hat on top space plus space 5 space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses space equals space 9

and the point (2, 1, 3).

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203. Find the equation of the plane passing through the line of intersection of the planes $straight r with rightwards arrow on top. space open parentheses straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top close parentheses space equals space 1 space space space and space space straight r with rightwards arrow on top. space space open parentheses 2 space straight i with hat on top space plus space 3 space straight j with hat on top space minus space straight k with hat on top close parentheses space plus space 4 space space equals 0$ and parallel to x-axis.

The equations of the planes are

straight r with rightwards arrow on top space space left parenthesis straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top right parenthesis space equals space 1 space space space and space straight r with rightwards arrow on top. space space left parenthesis 2 space straight i with hat on top space plus space 3 space straight j with hat on top space minus space space straight k with hat on top right parenthesis space plus space 4 space equals 0

left parenthesis straight x space straight i with hat on top space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top right parenthesis. space left parenthesis straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top right parenthesis space minus space 1 space equals space 0

and

open parentheses straight x space straight i with hat on top space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top close parentheses. space space open parentheses 2 space straight i with hat on top space plus space 3 space straight j with hat on top space minus straight k with hat on top close parentheses space plus space 4 space equals 0

straight x plus straight y plus straight z minus 1 space equals space 0

...(1)
and

2 straight x plus 3 straight y minus straight z plus 4 space equals space 0

...(2)
Any plane through the intersection of planes (1) and (2) is
(x + y + z – 1) + k (2x + 3y – r + 4) = 0 ...(3)
or (2k + 1) at + (3k + 1) y + (– k + 1) z + (4k – 1) = 0
Its direction ratios are 2k + 1, 3k + 1, – k + 1.
Since plane is parallel to x-axis with direction ratios 1, 0, 0.
∴ 1 (2k + 1) + (0) (3k + 1) + 0 (– k + 1) = 0

therefore space space space space space 2 straight k plus 1 space equals space 0 space space space space space space space space space rightwards double arrow space space space space 2 space straight k space equals space space minus 1 space space space space space space space space space rightwards double arrow space space space straight k space equals space minus 1 half

Putting

straight k equals negative 1 half space in space left parenthesis 3 right parenthesis comma space we space get comma

open parentheses straight x plus straight y plus straight z minus 1 close parentheses space minus space 1 half left parenthesis 2 straight x plus 3 straight y minus straight z plus 4 right parenthesis space equals space 0

2 space left parenthesis straight x plus straight y plus straight z minus 1 right parenthesis space minus space left parenthesis 2 straight x plus 3 straight y minus straight z plus 4 right parenthesis space equals space 0

or 2x + 2y + 2z – 2 – 2x – 3y + z – 4 = 0
or – y + 3z – 6 = 0
or y - 3z + 6 = 0
Which is required equation of plane.

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204. Find the equation of the plane through the line of intersection of the planes 3x – 4y + 5z = 10, 2x + 2y – 3z = 4 and parallel to the line x = 2y = 3z.

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205. Find the equation of the plane which contains the line of intersection of the planes

straight r with rightwards arrow on top. space open parentheses straight i with hat on top space plus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses space minus space 4 space equals 0 space space space and space space straight r with rightwards arrow on top. space space left parenthesis 2 straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top right parenthesis space plus space 5 space equals space 0

and which is perpendicular to the plane

straight r with rightwards arrow on top. space open parentheses 5 straight i with hat on top space plus space 3 straight j with hat on top space minus space 6 straight k with hat on top space close parentheses space plus space 8 space equals 0 space.

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Short Answer Type

206. Find the equation of plane passing through the line of intersection of the planes 2x – y = 0 and 3z – y = 0 and perpendicular to the plane 4x + 5y – 3Z = 8.

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207. Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.

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Long Answer Type

208. Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.

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209. Find the equation of the plane passing through the line of intersection of planes x + y – 2 z + 3 = 0 and 3 x – y – 2 z – 4 = 0 and perpendicular to the plane 2x + 3 y – z + 1 = 0.

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