Let  Find a vector  which is perpendicular to both  from M

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 Multiple Choice QuestionsLong Answer Type

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201.

Let straight a with rightwards arrow on top space equals straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top comma space space straight b with rightwards arrow on top space equals space 3 straight i with hat on top space minus space 2 straight j with hat on top space plus space 7 straight k with hat on top space and space straight c with rightwards arrow on top space equals space 2 straight i with hat on top minus straight j with hat on top minus 4 straight k with hat on top. Find a vector straight d with rightwards arrow on top which is perpendicular to both straight a with rightwards arrow on top space and space straight b with rightwards arrow on top space and space straight c with rightwards arrow on top. space straight d with rightwards arrow on top space equals space 15.


Here straight a with rightwards arrow on top space equals space straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top comma space space straight b with rightwards arrow on top space equals 3 straight i with hat on top minus 2 straight j with hat on top plus 7 straight k with hat on top comma space straight c with rightwards arrow on top space equals space 2 straight i with hat on top minus straight j with hat on top minus 4 straight k with hat on top
Let straight d with rightwards arrow on top space equals space straight d subscript 1 straight i with hat on top plus straight d subscript 2 straight j with hat on top plus straight d subscript 3 straight k with hat on top
Since straight d with rightwards arrow on top is perpendicular to straight a with rightwards arrow on top space and space straight b with rightwards arrow on top
therefore space space straight d with rightwards arrow on top. space straight a with rightwards arrow on top space equals 0 space space space and space straight d with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 0
therefore space space space space space straight d subscript 1 plus 4 straight d subscript 2 space plus space 2 straight d subscript 3 space equals space 0                                               ...(1)
and    3 space straight d subscript 1 space minus space 2 space straight d subscript 2 space plus space 7 space straight d subscript 3 space equals space 0                                      ...(2)
Again straight c with rightwards arrow on top. straight d with rightwards arrow on top space equals space 15
therefore space space space 2 space straight d subscript 1 space minus space straight d subscript 2 space plus space 4 space straight d subscript 3 space equals space 15                                         ...(3)
Multiplying (1) by 7 and (2) by -2, we get,
                  7 space straight d subscript 1 space plus space 28 space straight d subscript 2 space plus space 14 space straight d subscript 3 space equals space 0
minus 6 space straight d subscript 1 space plus space 4 space straight d subscript 2 space minus space 14 space straight d subscript 3 space equals space 0
Adding,                 straight d subscript 1 space plus space 32 space straight d subscript 2 space equals space 0                               ...(4)
Multiplying (1) by 2  and (3) by -1, we get,
               2 space straight d subscript 1 plus space 8 space straight d subscript 2 space plus space 4 space straight d subscript 3 space equals space 0
minus 2 space straight d subscript 1 plus space straight d subscript 2 space minus space 4 space straight d subscript 3 space equals space 0
Adding,          9 space straight d subscript 2 space equals space minus 15 comma space space space space space therefore space space straight d subscript 2 space equals space minus 15 over 9 space equals space minus 5 over 3
From (4),   straight d subscript 1 minus 160 over 3 space equals space 0 comma            therefore space space space straight d subscript 1 space equals space 160 over 3
From (1),    160 over 3 minus 20 over 3 plus 2 space straight d subscript 3 space equals space 0 comma space space space space space space therefore space space space 2 space straight d subscript 3 space equals space 20 over 3 minus 160 over 3

therefore space space 2 space straight d subscript 3 space equals space fraction numerator 20 minus 160 over denominator 3 end fraction space equals space minus 140 over 3 space space space space space space rightwards double arrow space space space space straight d subscript 3 space equals space minus 70 over 3

therefore space space required space vector space straight d with rightwards arrow on top space equals space 160 over 3 straight i with hat on top minus 5 over 3 straight j with hat on top space minus 70 over 3 straight k with hat on top
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 5 over 3 left parenthesis 32 space straight i with hat on top space minus space straight j with hat on top space minus space 14 space straight k with hat on top right parenthesis



 

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202.

Let straight a with rightwards arrow on top space equals straight i with hat on top space minus straight j with hat on top comma space straight b with rightwards arrow on top space equals space 3 straight j with hat on top space minus space straight k with hat on top space and space straight c with rightwards arrow on top space equals 7 straight i with hat on top space minus space straight k with hat on top. Find a vector straight d with rightwards arrow on top which is perpendicular to both straight a with rightwards arrow on top space and space straight b with rightwards arrow on top and straight c with rightwards arrow on top. straight d with rightwards arrow on top space equals space 1

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203. Dot product of a vector with vectors 2 straight i with hat on top space plus space 3 straight j with hat on top space plus space straight k with hat on top comma space space 4 straight i with hat on top space plus space straight j with hat on top space and space straight i with hat on top space minus space 3 straight j with hat on top space minus space 7 straight k with hat on top are respectively 9, 7 and 6. Find the vector.
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 Multiple Choice QuestionsShort Answer Type

204. Dot product of a vector with vectors 3 space straight i with hat on top minus 5 space straight k with hat on top comma space space space 2 straight i with hat on top plus space 7 space straight j with hat on top space and space straight i with hat on top plus straight j with hat on top plus straight k with hat on top are respectively – 1, 6 and 5. Find the vector.
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 Multiple Choice QuestionsLong Answer Type

205. Find the scalar components of a unit vector which is perpendicular to the vectors straight i with hat on top plus 2 space straight j with hat on top space minus space straight k with hat on top space space and space space 3 straight i with hat on top minus straight j with hat on top plus 2 space straight k with hat on top.
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 Multiple Choice QuestionsShort Answer Type

206.

If straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top space equals space 0 with rightwards arrow on top comma show that the angle straight theta between the vectors straight b with rightwards arrow on top space and space straight c with rightwards arrow on top is given by
cos space straight theta space equals space fraction numerator straight a squared minus straight b squared minus straight c squared over denominator 2 space open vertical bar straight b with rightwards arrow on top close vertical bar space open vertical bar straight c with rightwards arrow on top close vertical bar end fraction

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207.

If straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top space equals 0 with rightwards arrow on top space and space open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 3 comma space space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 5 comma space space open vertical bar straight c with rightwards arrow on top close vertical bar space equals 7 comma show that the angle between straight a with rightwards arrow on top space and space straight b with rightwards arrow on top is 60°.

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208.

Let straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top be three vectors of magnitude 5, 3, 1 respectively. If each one is perpendicular to the sum of other two vectors, prove that open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar space equals space square root of 35.

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209.

If 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 are three mutually perpendicular vectors of equal magnitude find the angle between straight a with rightwards arrow on top space and space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis.

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 Multiple Choice QuestionsLong Answer Type

210.

If 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 are mutually perpendicular vectors of equal magnitude, show that they are equally inclined to the vector left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis

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