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Vector Algebra

Question
CBSEENMA12034127

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.

Solution

Let straight r with rightwards arrow on top space equals space straight x straight i with hat on top space plus space straight y straight j with hat on top space plus space straight z straight k with hat on top be required vector. 
 From given conditions,
left parenthesis straight x straight i with hat on top space plus space straight y straight j with hat on top plus straight z straight k with hat on top right parenthesis. space left parenthesis 2 straight j with hat on top plus 3 straight j with hat on top plus straight k with hat on top right parenthesis space equals space 9 space space space rightwards double arrow space space space 2 straight x plus 3 straight y plus straight z space equals space 9               ...(1)
left parenthesis straight x straight i with hat on top space plus space straight y straight j with hat on top space plus straight z straight k with hat on top right parenthesis. space left parenthesis 4 straight i with hat on top plus straight j with hat on top right parenthesis space equals space 7 space space space space space space space rightwards double arrow space space space 4 straight x plus straight y space equals 7                             ...(2)
left parenthesis straight x straight i with hat on top plus space straight y straight j with hat on top space plus space straight z straight k with hat on top right parenthesis. space left parenthesis straight i with hat on top minus 3 straight j with hat on top minus 7 straight k with hat on top right parenthesis space equals 6 space space space space rightwards double arrow space space straight x minus 3 straight y minus 7 straight z space equals space 6                ...(3)
Multiplying (1) by 7 and (3) by 1, we get,
14x + 21 y + 7z = 63
x – 3y – 7z = 6
Adding, we get.
15x +18y = 69 or 5x + 6y = 23    …(4)
Multiplying (2) by 6 and (4) by 1, we get,
24x + 6y = 42    ...(5)
5x + 6y = 23    ...(6)
Subtracting (6) from (5), we get,
19x = 19 or x = 1
Putting x = 1 in (2), we get,
4 + y = 7 or y = 3
Putting x = 1, y = 3 in (1), we get,
2 + 9 + z = 9 or z = – 2
therefore space space space space straight r with rightwards arrow on top space equals space straight i with hat on top space plus space 3 straight j with hat on top minus 2 straight k with hat on top space is space required space vector.