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If A = 2i + j - 3k and B = i - 2j + k,find a vector of magnitude 5 perpendicular to both A and B. If A = 2i + j - 3k and B = i - 2j + k,find a vector of magnitude 5 perpendicular to both A and B.
Hi Let us suppose the vector is xi + yj + zk . Now take its dot product with each vector individually and equate it to zero , becoz since it is perpendicular to the given vectors , its dot product with the vectors must be zero. So, u get two eq from this , nw since there are three variables u need one more eq, so u use the magnitude of the vector as the third eq and solve these three eq for the unknowns.
Hi
Let us suppose the vector is xi + yj + zk . Now take its dot product with each vector individually and equate it to zero , becoz since it is perpendicular to the given vectors , its dot product with the vectors must be zero.
So, u get two eq from this , nw since there are three variables u need one more eq, so u use the magnitude of the vector as the third eq and solve these three eq for the unknowns.
find the cross product of the two vectors A and B,that will be perpendicular to both these vectors(call that vector as C).find the unit vector along the direction of vector C and mutiply with 5,u will get the reqiured answer.
$ - means squre root. the vector perpendicular to A & B is A x B A x B = i(1-6)-j(2+3)+k(-4-1) = -5(i+j+k) unit vector along A x B is =A x B / |A x B| = -5(i+j+k) / $(75) = (i + j + k )/ $3 therefore vector of magnitude 5 in directon of A x B is = 5(i + j + k)/ $3 and it may be in opposite direction that of A x B Ans.(+ or -)5(i+j+k)/$3
$ - means squre root.
the vector perpendicular to A & B is A x B
A x B = i(1-6)-j(2+3)+k(-4-1)
= -5(i+j+k)
unit vector along A x B is =A x B / |A x B|
= -5(i+j+k) / $(75)
= (i + j + k )/ $3
therefore vector of magnitude 5 in directon of A x B is = 5(i + j + k)/ $3
and it may be in opposite direction that of A x B
Ans.(+ or -)5(i+j+k)/$3
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