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Vector Algebra

Question
CBSEENMA12033987

If PQ with rightwards arrow on top space equals space 3 straight i with hat on top space plus space 2 straight j with hat on top space minus space straight k with hat on top and the coordinates of P are (1, – 1, – 2), find the coordinates of Q.

Solution

Here P = (1, -1, -2)    rightwards double arrow   P.V. of P = straight i with hat on top space minus space straight j with hat on top space minus space 2 straight k with hat on top
Let Q be left parenthesis straight lambda subscript 1 comma space straight lambda subscript 2 comma space straight lambda subscript 3 right parenthesis
therefore space space space space straight P. straight V. space of space straight Q space equals space straight lambda subscript 1 straight i with hat on top space space plus space straight lambda subscript 2 straight j with hat on top space plus space straight lambda subscript 3 straight k with hat on top
Now,  PQ with rightwards arrow on top space equals space straight P. straight V. space of space straight Q space minus space straight P. straight V. space of space straight P
rightwards double arrow space space space space space 3 straight i with hat on top plus space 2 straight j with hat on top space minus space straight k with hat on top space equals space left parenthesis straight lambda subscript 1 straight i with hat on top space plus space straight lambda subscript 2 straight j with hat on top space plus space straight lambda subscript 3 straight k with hat on top right parenthesis space minus space left parenthesis 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
rightwards double arrow space space space space 3 space straight i with hat on top space plus space 2 space straight j with hat on top space minus space straight k with hat on top space equals space left parenthesis straight lambda subscript 1 space minus space 1 right parenthesis straight i with hat on top space plus space left parenthesis straight lambda subscript 2 space plus space 1 right parenthesis space straight j with hat on top space plus space left parenthesis straight lambda subscript 3 space plus 2 right parenthesis space straight k with hat on top
rightwards double arrow space space space 3 space equals space straight lambda subscript 1 space minus space 1 comma space space 2 space equals space straight lambda subscript 2 plus 1 comma space space minus 1 space equals space straight lambda subscript 3 plus 2
rightwards double arrow space space space straight lambda subscript 1 space equals space 4 comma space space space space space straight lambda subscript 2 space equals space 1 comma space space space straight lambda subscript 3 space equals space minus 3 comma
therefore space space space space straight Q space is space left parenthesis 4 comma space 1 comma space minus 3 right parenthesis.