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Three Dimensional Geometry

Short Answer Type

241. Find the distance between the point P(6, 5, 9) and the plane determined by the points A(3,–1, 2), B(5, 2,4) and C(–1,–1, 6).

The equation of plane through A(3, –1, 2), B(5, 2, 4), C(–1, –1, 6) is

open vertical bar table row cell straight x minus 3 end cell cell straight y plus 1 end cell cell straight z minus 2 end cell row cell 5 minus 3 end cell cell 2 plus 1 end cell cell 4 minus 2 end cell row cell negative 1 minus 3 end cell cell negative 1 plus 1 end cell cell 6 minus 2 end cell end table close vertical bar space equals space 0

open vertical bar table row cell straight x minus 3 end cell cell straight y plus 1 end cell cell straight z minus 2 end cell row 2 3 2 row cell negative 4 end cell 0 4 end table close vertical bar space equals space 0

or space space left parenthesis straight x minus 3 right parenthesis space open vertical bar table row 3 2 row 0 4 end table close vertical bar space minus space left parenthesis straight y plus 1 right parenthesis space open vertical bar table row 2 2 row cell negative 4 end cell 4 end table close vertical bar space plus space left parenthesis straight z minus 2 right parenthesis space open vertical bar table row 2 3 row cell negative 4 end cell 0 end table close vertical bar space equals space 0 or space space left parenthesis straight x minus 3 right parenthesis thin space left parenthesis 12 minus 0 right parenthesis space minus space left parenthesis straight y plus 1 right parenthesis space left parenthesis 8 plus 8 right parenthesis space plus space left parenthesis straight z minus 2 right parenthesis thin space left parenthesis 0 plus 12 right parenthesis space equals space 0 or space space 12 left parenthesis straight x minus 3 right parenthesis space minus space 16 space left parenthesis straight y plus 1 right parenthesis space plus 12 space left parenthesis straight z minus 2 right parenthesis space equals space 0 or space space 3 space left parenthesis straight x minus 3 right parenthesis space minus space 4 space left parenthesis straight y plus 1 right parenthesis space plus space 3 space left parenthesis straight z minus 2 right parenthesis space equals space 0 or space 3 straight x minus 9 space minus space 4 straight y space minus 4 space plus space 3 straight z space minus space 6 space equals space space 0 or space space 3 straight x minus 4 straight y plus 3 straight z minus 19 space equals space 0

Let p be distance between P(6, 5, 9) and plane.

therefore space space space space straight p space equals space fraction numerator 3 space left parenthesis 6 right parenthesis space minus space 4 space left parenthesis 5 right parenthesis space plus space 3 space left parenthesis 9 right parenthesis space minus space 19 over denominator square root of left parenthesis 3 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared plus left parenthesis 3 right parenthesis squared end root end fraction space equals space fraction numerator 18 minus 20 plus 27 minus 19 over denominator square root of 9 plus 16 plus 9 end root end fraction

equals space fraction numerator 6 over denominator square root of 34 end fraction space equals space fraction numerator 6 square root of 34 over denominator 34 end fraction space equals space fraction numerator 3 square root of 34 over denominator 17 end fraction

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242. Find the distance of the point (2, 3, 4) from the plane

straight r with rightwards arrow on top. space open parentheses 3 space straight i with hat on top space minus space 6 space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses space equals space minus 11.

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243. Find the distance of a point (2, 5, -3) from the plane

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

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244. If the product of distances of the point (1,1,1) from the origin and the plane x – y + z + k = 0 be 5, then find the value of k.

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245. If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane

straight r with rightwards arrow on top. space left parenthesis 3 space straight i with hat on top space plus space 4 space straight j with hat on top space minus space 1 space 2 space straight k with hat on top right parenthesis space plus space 13 space equals space 0 comma

then find value of p.

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246.

Find the angle between the line
$straight r with rightwards arrow on top space equals space left parenthesis 2 space straight i with hat on top space plus space 2 space straight j with hat on top space plus space straight k with hat on top right parenthesis space plus space straight lambda space left parenthesis 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis$ and the plane
$straight r with rightwards arrow on top. space left parenthesis 3 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 5 space straight k with hat on top right parenthesis space equals space 4.$

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247.

Find the angle between the line $straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space plus 3 space straight j with hat on top space plus space 9 space straight k with hat on top space plus space straight lambda left parenthesis 2 straight i with hat on top space plus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top right parenthesis$ and the plane $straight r with rightwards arrow on top. space open parentheses straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top close parentheses space equals space 5$

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248.

Find the angle between the line $straight r with rightwards arrow on top space equals space open parentheses 2 space straight i with hat on top space minus straight j with hat on top space plus space 3 space straight k with hat on top close parentheses space plus space straight lambda open parentheses 3 space straight i with hat on top space minus space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses$ and the plane $straight r with rightwards arrow on top. space open parentheses straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top close parentheses space equals space 3$

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249. Find the angle between the line

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 minus space straight k with hat on top close parentheses space space plus space straight lambda space open parentheses straight i with hat on top space minus space straight j with hat on top space plus space straight k with hat on top close parentheses

and the plane

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

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250.

Find the angle between the line $straight x over 4 space equals space fraction numerator straight y minus 2 over denominator 3 end fraction space equals space fraction numerator straight z minus 1 over denominator 6 end fraction$ and the plane x + 2y + 2 + 3 = 0

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