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Vector Algebra

Question
CBSEENMA12034126

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

Solution

Here straight a with rightwards arrow on top space equals straight i with hat on top space minus space 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 comma space space straight c with rightwards arrow on top space equals space 7 straight i with hat on top space minus space straight k with hat on top
Let straight d with rightwards arrow on top space equals straight d subscript 1 straight i with hat on top space plus space straight d subscript 2 straight j with hat on top space plus space 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 space space straight d with rightwards arrow on top. space straight a with rightwards arrow on top space equals space 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 space minus space straight d subscript 2 space equals space 0 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 space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
and     3d2 – d3 = 0    ...(2)
Also,         straight c with rightwards arrow on top. space straight d with rightwards arrow on top space equals space 1 space space space space space space space space space rightwards double arrow space space space space 7 space straight d subscript 1 space minus space straight d subscript 3 space equals space 1               ...(3)

Multiplying (1) by 3 and (2) by 1, we get,
3d1 – 3d2 = 0
3d2 – d3 = 0
Adding, 3d1 – d3 = 0    ...(4)
Subtracting (4) from (3), we get,
                           4 straight d subscript 1 space equals space 1                         rightwards double arrow space space space straight d subscript 1 space equals 1 fourth
From (1),  1 fourth minus straight d subscript 2 space equals space 0                             rightwards double arrow space space space straight d subscript 2 space equals space 1 fourth
From (2),  3 over 4 minus straight d subscript 3 space equals space 0 space space space space space space space rightwards double arrow space space space space straight d subscript 3 space equals space 3 over 4
therefore space space straight d with rightwards arrow on top space equals space 1 fourth straight i with hat on top space plus 1 fourth straight j with hat on top space plus space 3 over 4 straight k with hat on top
therefore space space space straight d with rightwards arrow on top space equals space 1 fourth left parenthesis straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top right parenthesis