In the following cases, determine whether the given planes are p

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 Multiple Choice QuestionsShort Answer Type

181. Find the vector equation of the line through the origin which is perpendicular to the plane straight r with rightwards arrow on top. space open parentheses straight i with hat on top space minus space 2 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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182. Find the vector and cartesian equations of the planes:
that passes through the point (1, 0, – 2) and the normal to the plane is straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top.
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183. Find the vector and cartesian equations of the planes:
that passes through the point (1, 4, 6) and the normal vector to the plane is straight i with hat on top space minus space 2 space straight j with hat on top space plus space straight k with hat on top.
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 Multiple Choice QuestionsLong Answer Type

184. Find the vector equation of the line passing through the point (3, 1, 2) and perpendicular to the plane straight r with rightwards arrow on top. space open parentheses 2 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. Find also the point of intersection of this line and plane. 
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 Multiple Choice QuestionsShort Answer Type

185. Find the angle between the two planes
3 x – 6 y + 2 z = 7 and 2 x + 2 y – 2 z = 5.
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186. Find the angle between the two planes
2 x + y – 2 z = 5 and 3x – 6 y – 2 z = 7 using vector method.
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187. Find the angle between the planes whose vector equations are
straight r with rightwards arrow on top. space open parentheses 2 straight i with hat on top space plus space 2 straight j with hat on top space minus space 3 straight k with hat on top close parentheses space equals space 5 space space space and space straight r with rightwards arrow on top. space open parentheses 3 straight i with hat on top space minus space 3 straight j with hat on top space plus space 5 straight k with hat on top close parentheses space equals space 3.
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188. Find the angle between the planes whose vector equations are
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 minus space 2 space straight k with hat on top close parentheses space equals space 3 space space and space space space straight r with rightwards arrow on top. space space open parentheses 2 straight i with hat on top space minus space 2 straight j with hat on top space plus space straight k with hat on top close parentheses space equals space 2
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189. Find the angle between the planes
straight r with rightwards arrow on top. space open parentheses 3 space straight i with hat on top space minus space 4 space straight j with hat on top space plus space 5 space straight k with hat on top close parentheses space equals space 0 space space and space straight r with rightwards arrow on top. space space left parenthesis 2 straight i with hat on top space minus space straight j with hat on top space minus space 2 space straight k with hat on top right parenthesis space equals space 0

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190. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them:
7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0


The equations of given planes are
7x + 5y + 6z + 30 = 0    ...(1)
and 3x – y – 10z + 4 = 0    ...(2)
Direction ratios of normal to plane (1) are 7, 5, 6
Direction ratios of normal to plane (2) are 3, –1, –10.
Now 7 over 3 not equal to fraction numerator 5 over denominator negative 1 end fraction not equal to fraction numerator 6 over denominator negative 10 end fraction
∴    planes (1) and (2) are not parallel.
Again (7) (3) + (5) (– 1) + (6) (– 10) = 21 – 5 – 60 = – 44 ≠ 0
∴  planes (1) and (2) are not perpendicular to each other.
Let θ be angle between planes (1) and (2).
therefore space space space space space space cos space straight theta space equals space fraction numerator open vertical bar left parenthesis 7 right parenthesis thin space left parenthesis 3 right parenthesis space plus space left parenthesis 5 right parenthesis thin space left parenthesis negative 1 right parenthesis space plus space left parenthesis 6 right parenthesis thin space left parenthesis negative 10 right parenthesis close vertical bar over denominator square root of left parenthesis 7 right parenthesis squared plus left parenthesis 5 right parenthesis squared plus left parenthesis 6 right parenthesis squared end root space square root of left parenthesis 3 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared plus left parenthesis negative 10 right parenthesis squared end root end fraction
                     equals space fraction numerator open vertical bar 21 minus 5 minus 60 close vertical bar over denominator square root of 49 plus 25 plus 36 end root space square root of 9 plus 1 plus 100 end root end fraction space equals space fraction numerator open vertical bar negative 44 close vertical bar over denominator square root of 110 space square root of 110 end fraction
space equals 44 over 110 space equals 22 over 55 space equals 2 over 5
therefore space space space space space space straight theta space equals space cos to the power of negative 1 end exponent open parentheses 2 over 5 close parentheses

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