Find the vector equation of the plane passing through three poin

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 Multiple Choice QuestionsShort Answer Type

321.

Find the coordinates of the point, where the line fraction numerator straight x minus 2 over denominator 3 end fraction equals fraction numerator straight y plus 1 over denominator 4 end fraction equals fraction numerator straight z minus 2 over denominator 2 end fraction intersects the plane straight x minus straight y plus straight z minus 5 space equals space 0. Also find the angle between the line and the plane. 

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322.

Find the vector equation of the plane which contains the line of intersection of the planes. straight r with rightwards arrow on top. open parentheses straight i with hat on top plus 2 straight j with hat on top plus 3 straight k with hat on top close parentheses minus 4 space equals 0 and straight r with rightwards arrow on top. open parentheses 2 straight i with hat on top plus straight j with hat on top minus straight k with hat on top close parentheses plus 5 space equals 0 and which is perpendicular to the plane straight r with rightwards arrow on top. space open parentheses 5 straight i with hat on top plus 3 straight j with hat on top minus 6 straight k with hat on top close parentheses plus 8 space equals space 0

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 Multiple Choice QuestionsLong Answer Type

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323.

Find the vector equation of the plane passing through three points with position vectors straight i with hat on top plus straight j with hat on top minus 2 straight k with hat on top comma space 2 straight i with hat on top minus straight j with hat on top plus straight k with hat on top space and space straight i with hat on top plus 2 straight j with hat on top plus straight k with hat on top. Also, find the coordinates of the point of intersection of this plane and the line straight r with rightwards arrow on top space equals 3 straight i with hat on top minus straight j with hat on top minus straight k with hat on top plus straight lambda open parentheses 2 straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top close parentheses.


Let the position vectors of the three points be, 
straight a with rightwards arrow on top equals straight i with hat on top plus straight j with hat on top minus 2 straight k with hat on top comma space space straight b with rightwards arrow on top equals 2 straight i with hat on top minus straight j with hat on top plus straight k with hat on top space space and space straight c with rightwards arrow on top equals straight i with hat on top plus 2 straight j with hat on top plus straight k with hat on top.
So, the equation of the plane passing through the point straight a with rightwards arrow on top comma space straight b with rightwards arrow on top space and space straight c with rightwards arrow on top space is
open parentheses straight r with rightwards arrow on top minus straight a with rightwards arrow on top close parentheses. open square brackets left parenthesis straight b with rightwards arrow on top. negative straight c with rightwards arrow on top right parenthesis space cross times space left parenthesis straight c with rightwards arrow on top minus straight a with rightwards arrow on top right parenthesis close square brackets space equals space 0
rightwards double arrow space space open square brackets straight r with rightwards arrow on top. open parentheses straight i with hat on top plus straight j with hat on top minus 2 straight k with hat on top close parentheses close square brackets. space open square brackets left parenthesis straight i with hat on top minus 3 straight j with hat on top right parenthesis cross times left parenthesis straight j with hat on top plus 3 straight k with hat on top right parenthesis close square brackets equals 0
rightwards double arrow space open square brackets straight r with rightwards arrow on top minus open parentheses straight i with hat on top plus straight j with hat on top minus 2 straight k with hat on top close parentheses close square brackets. space open parentheses straight k with hat on top minus 3 straight j with hat on top minus 9 straight i with hat on top close parentheses equals 0
rightwards double arrow straight r with rightwards arrow on top. left parenthesis negative 9 straight i with hat on top minus 3 straight j with hat on top plus straight k with hat on top right parenthesis plus 14 space equals space 0
rightwards double arrow space straight r with rightwards arrow on top. open parentheses 9 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top close parentheses equals 14 space space space space space space space space space... left parenthesis 1 right parenthesis
So, the vector equation of the required plane is straight r with rightwards arrow on top. space open parentheses 9 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top close parentheses space equals space 14.
The equation of the given line is straight r with rightwards arrow on top space equals space open parentheses 3 straight i with hat on top minus straight j with hat on top minus straight k with hat on top close parentheses plus straight lambda left parenthesis 2 straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top right parenthesis.
Position vector of any point on the given line is

straight r with rightwards arrow on top space equals space left parenthesis 3 plus 2 straight lambda right parenthesis straight i with hat on top plus left parenthesis negative 1 minus 2 straight lambda right parenthesis straight j with hat on top plus left parenthesis negative 1 plus straight lambda right parenthesis straight k with hat on top   ....(2)
The point (2) lies on plane (1) if, 

open square brackets left parenthesis 3 plus 2 straight lambda right parenthesis straight i with hat on top plus left parenthesis negative 1 minus 2 straight lambda right parenthesis straight j with hat on top plus left parenthesis negative 1 plus straight lambda right parenthesis straight k with hat on top close square brackets. left parenthesis 9 straight i with hat on top plus 3 straight j with hat on top minus straight k with hat on top right parenthesis space equals space 14
rightwards double arrow space space 9 left parenthesis 3 plus 2 straight lambda right parenthesis plus 3 left parenthesis negative 1 minus 2 straight lambda right parenthesis minus left parenthesis negative 1 plus straight lambda right parenthesis equals space 14
rightwards double arrow space space 11 straight lambda plus 25 space equals 14
rightwards double arrow space space straight lambda space equals negative 1
Putting straight lambda equals negative 1 in (2), we have
straight r with rightwards arrow on top equals left parenthesis 3 plus 2 straight lambda right parenthesis straight i with hat on top plus left parenthesis negative 1 minus 2 straight lambda right parenthesis straight j with hat on top plus left parenthesis negative 1 plus left parenthesis negative 1 right parenthesis right parenthesis straight k with hat on top
space space equals open parentheses 3 plus 2 left parenthesis negative 1 right parenthesis close parentheses straight i with hat on top plus left parenthesis negative 1 minus 2 left parenthesis negative 1 right parenthesis right parenthesis straight j with hat on top plus left parenthesis negative 1 plus left parenthesis negative 1 right parenthesis right parenthesis straight k with hat on top
equals straight i with hat on top plus straight j with hat on top minus 2 straight k with hat on top
Thus, the position vector of the point of intersection of the given line and plane (1) is 

straight i with hat on top plus straight j with hat on top minus 2 straight k with hat on top and its co-ordinates are (1, 1, -2).

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324.

If straight a with rightwards arrow on top comma straight b with rightwards arrow on top comma space and space straight c with rightwards arrow on top are mutually perpendicular vectors of equal magnitudes, show that the vector stack straight a space with rightwards arrow on top space plus straight b with rightwards arrow on top space plus straight c with rightwards arrow on top is equally straight a with rightwards arrow on top comma straight b with rightwards arrow on top comma space and space straight c with rightwards arrow on top inclined to Also, find the angle which stack straight a space with rightwards arrow on top space plus straight b with rightwards arrow on top space plus straight c with rightwards arrow on top with straight a with rightwards arrow on top comma space straight b with rightwards arrow on top space or space straight c with rightwards arrow on top.

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325. Let space straight a with rightwards arrow on top space equals space straight i with hat on top space plus straight j with hat on top space plus stack straight k comma with hat on top space
straight b with rightwards arrow on top space equals space straight i with hat on top space and
straight c with rightwards arrow on top space equals space straight c subscript 1 straight i with hat on top space plus straight c subscript 2 straight j with hat on top space plus straight c subscript 3 straight k with hat on top space space then
a) Let c1 = 1 and c2 = 2, find c3 which makes straight a with rightwards arrow on top space comma straight b with rightwards arrow on top space and space straight c with rightwards arrow on top coplanar.
b) If c2 = –1 and c3 = 1, show that no value of c1 can make straight a with rightwards arrow on top space comma straight b with rightwards arrow on top space and space straight c with rightwards arrow on top coplanar.
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 Multiple Choice QuestionsShort Answer Type

326.

Find the magnitude of each of the two vectors a and b, having the same magnitude such that the angle between them is 60o and their scalar product is 9/2.


327.

If θ is the angle between two vectors i^ - 2j^ + 3k^ and 3i^ - 2j^ + k^ find sin θ.


328.

Let a = 4i^ + 5j^ - k^ , b = i^ - 4j^ + 5k^ and c = 3i^ + j^- k^. Find a vector d which perpendicular to both  c and b and d. a = 21


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329.

Find a unit vector in the direction of  a = 3i ^ - 2j^ + 6k


330.

Find the angle between the vectors a = i^ - j^ + k^    and   b = 1^ + j^ - k^ 


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