Find the vector and cartesian form of the equation of the plane

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 Multiple Choice QuestionsShort Answer Type

271. Find the distance of a point P (5, 3, 4) from the point where the line fraction numerator straight x minus 3 over denominator 1 end fraction space equals fraction numerator straight y minus 4 over denominator 2 end fraction space equals fraction numerator straight z minus 5 over denominator 2 end fraction meets the plane x + y + z = 2.
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272. Find the distance of a point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line straight x over 2 space equals space straight y over 3 space equals space fraction numerator straight z over denominator negative 6 end fraction
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273. Find the distance of the point (- 2, 3, - 4) from the line fraction numerator straight x plus 2 over denominator 3 end fraction space equals space fraction numerator 2 straight y plus 3 over denominator 4 end fraction space equals space fraction numerator 3 straight z plus 4 over denominator 5 end fraction measured  parallel to the plane 4 x + 12 y - 3z + 1 = 0.
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 Multiple Choice QuestionsLong Answer Type

274. Find the distance of the point (–1, –5, –10) from the point of intersection of the line straight r with rightwards arrow on top space equals space 2 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 space plus space straight lambda space open parentheses 3 straight i with hat on top space plus space 4 space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses space and space the space plane space straight r with rightwards arrow on top. space open parentheses straight i with hat on top minus straight j with hat on top space plus space space straight k with hat on top close parentheses space equals space 5.
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 Multiple Choice QuestionsShort Answer Type

275. Find the distance between the point with position vector negative straight i with hat on top space minus space 5 space straight j with hat on top space minus space 10 space straight k with hat on top and the point of intersection of the line fraction numerator straight x minus 2 over denominator 3 end fraction space equals space fraction numerator straight y plus 1 over denominator 4 end fraction space equals space fraction numerator straight z minus 2 over denominator 12 end fraction with the plane x – y + z = 5.
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276.

Find the distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0, measured parallel to the line fraction numerator straight x plus 3 over denominator 3 end fraction space equals space fraction numerator straight y minus 2 over denominator 6 end fraction space equals space straight z over 2

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277. Determine whether the lines:
         straight r with rightwards arrow on top space equals space straight i with hat on top space minus space straight j with hat on top space plus space straight lambda open parentheses 2 straight i with hat on top space plus space straight k with hat on top close parentheses
and   straight r with rightwards arrow on top space equals space 2 straight i with hat on top space minus space straight j with hat on top space plus space straight mu space open parentheses straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top close parentheses space space intersect.
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278. Show that the lines
straight r with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top space plus space straight lambda open parentheses 3 space straight i with hat on top space minus space straight j with hat on top close parentheses space space and space straight r with rightwards arrow on top space equals space 4 space straight i with hat on top space minus space straight k with hat on top space plus space straight mu space open parentheses 2 straight i with hat on top space plus space 3 space straight k with hat on top close parentheses intersect.
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 Multiple Choice QuestionsLong Answer Type

279. Find the equation of plane containing the coplanar lines
straight r with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top 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 space space and space straight r with rightwards arrow on top space equals space 4 space straight i with hat on top space plus space 2 straight k with hat on top space plus space straight mu open parentheses 2 space straight i with hat on top space minus space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses.
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 Multiple Choice QuestionsShort Answer Type

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280. Find the vector and cartesian form of the equation of the plane containing two lines:
            straight r with rightwards arrow on top space equals straight i with hat on top space plus space 2 space straight j with hat on top space minus space 4 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 6 space straight k with hat on top right parenthesis
and       straight r with rightwards arrow on top space equals space 3 straight i with hat on top space plus space 3 space straight j with hat on top space minus space 5 space straight k with hat on top space plus space straight mu space left parenthesis negative 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 8 space straight k with hat on top right parenthesis


Here,   straight a with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 4 space straight k with hat on top comma space space straight b with rightwards arrow on top space equals space 2 straight i with hat on top space plus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top comma space space straight c with rightwards arrow on top space equals space minus 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 8 space straight k with hat on top
The vector equation of plane is
             left parenthesis straight r with rightwards arrow on top space minus space straight a with rightwards arrow on top right parenthesis. space space open parentheses straight b with rightwards arrow on top space cross times straight c with rightwards arrow on top close parentheses space equals space 0
or     open square brackets straight r with rightwards arrow on top space minus space left parenthesis straight i with hat on top space plus space 2 space straight j with hat on top space minus space 4 space straight k with hat on top right parenthesis close square brackets. space open square brackets left parenthesis 2 straight i with hat on top space plus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top right parenthesis space cross times space space left parenthesis negative 2 space straight i with hat on top space plus space space 3 space straight j with hat on top space plus space space 8 space straight k with hat on top close square brackets space equals space 0
Cartesian equation of plane is
                       open vertical bar table row cell straight x minus 1 end cell cell straight y minus 2 end cell cell straight z plus 4 end cell row 2 3 6 row cell negative 2 end cell 3 8 end table close vertical bar space equals space 0
or space space left parenthesis straight x minus 1 right parenthesis thin space open vertical bar table row 3 6 row 3 8 end table close vertical bar space minus space left parenthesis straight y minus 2 right parenthesis space open vertical bar table row 2 6 row cell negative 2 end cell 8 end table close vertical bar space plus space left parenthesis straight z plus 4 right parenthesis space open vertical bar table row 2 3 row cell negative 2 end cell 3 end table close vertical bar space equals space 0
or    (x – 1) (24 – 18) – (y – 2) (16 + 12) + (z + 4) (6 + 6) = 0
or    6 (x – 1) – 28 (y – 2) + 12 (z + 4) = 0
or    3 (x – 1) – 14 (y – 2) + 6 (z + 4) = 0
or    3 x – 3 – l4y + 28 + 6z + 24 = 0
or    3 x – 14 y + 6 z + 49 = 0

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