If the product of distances of the point (1,1,1) from the origin

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 Multiple Choice QuestionsShort 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).
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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.


Let d1 , d2 be the distances of the point (1, 1, 1) from the origin (0, 0, 0) and plane x – y + z + k = 0.
therefore space space space space space space space straight d subscript 1 space equals space square root of left parenthesis 1 minus 0 right parenthesis squared plus left parenthesis 1 minus 0 right parenthesis squared plus left parenthesis 1 minus 0 right parenthesis squared end root space equals space square root of 1 plus 1 plus 1 end root space equals space square root of 3
and space space space straight d subscript 2 space equals space fraction numerator open vertical bar 1 minus 1 plus 1 plus straight k close vertical bar over denominator square root of 1 plus 1 plus 1 end root end fraction space equals space fraction numerator open vertical bar 1 plus straight k close vertical bar over denominator square root of 3 end fraction
From the given condition,
                   straight d subscript 1 space straight d subscript 2 space equals space 5

therefore space space space space square root of 3 space cross times fraction numerator open vertical bar 1 plus straight k close vertical bar over denominator square root of 3 end fraction space equals space 5 space space space space space space rightwards double arrow space space space space open vertical bar 1 plus straight k close vertical bar space equals space 5 space space space space space rightwards double arrow space space space space 1 plus straight k space equals space plus-or-minus 5
therefore space space space space space space straight k space equals space minus 6 comma space space 4.
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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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