Find the vector equation of the line passing through the point (

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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

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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. 


The equation of plane is 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                      ....(1)
Vector along the normal to the plane is 2 straight i with hat on top minus straight j with hat on top plus straight k with hat on top
Now required line passes through point (3, 1, 2) with position vector 3 space straight i with hat on top space plus space straight j with hat on top space plus space 2 space straight k with hat on top and is parallel to the vector 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.
                                straight r with rightwards arrow on top space equals space open parentheses 3 space straight i with hat on top space plus space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses space plus straight lambda 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              ...(2)
        ∴ equation of line is                                                         open square brackets because space straight r with rightwards arrow on top space equals straight a with rightwards arrow on top space plus straight lambda straight m with rightwards arrow on top close square brackets
Now line (2) meets plane (1) when
        open curly brackets 3 space straight i with hat on top space plus space straight j with hat on top space plus space 2 space straight k with hat on top space plus space straight lambda space left parenthesis 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 right parenthesis close curly brackets space. space left parenthesis 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 right parenthesis space equals space 4
straight i. straight e. space space space when space open curly brackets left parenthesis 3 plus 2 straight lambda right parenthesis straight i with hat on top space plus space left parenthesis 1 minus straight lambda right parenthesis space straight j with hat on top space plus space left parenthesis 2 plus straight lambda right parenthesis space straight k with hat on top close curly brackets. 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
straight i. straight e. space when space space 2 space left parenthesis 3 plus 2 space straight lambda right parenthesis space plus space left parenthesis negative 1 right parenthesis thin space left parenthesis 1 minus straight lambda right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis 2 plus straight lambda right parenthesis space equals space 4
straight i. straight e. space when space 6 plus 4 straight lambda minus 1 plus straight lambda plus 2 plus straight lambda space equals space 4
straight i. straight e. space when space 6 straight lambda space equals space minus 3 space comma space space space space space straight i. straight e. comma space when space space straight lambda space space equals space minus 1 half
Putting this value of λ in (2), we get the position vector of the point of intersection of (1) and (2) as
                    straight r with rightwards arrow on top space equals space open parentheses 3 space straight i with hat on top space plus space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses space minus space 1 half left parenthesis 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 right parenthesis space equals space 2 space straight i with hat on top space plus space 3 over 2 straight j with hat on top space plus 3 over 2 straight k with hat on top
therefore space space space required space point space is space open parentheses 2 comma space 3 over 2 comma space 3 over 2 close parentheses.
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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

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