24a=3i+2j+6k
b=2i+j+k
c=i+j-k
Actually explaianation was given
r=ax(bxc)
plz explain this
Find vector which is orthogonal to vector 3i+2j+6k and is coplanar with 2i+j+k and i+j-k
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4 Answers
62k(2i+j+k)+m(i+j-k) as the vector coplanar to the two vectors.. then use the condition that it is orthogonal with vector 2i+2j+6k
by taking dot product..
so (2k+m).2+(k+m).2+(k-m).6=0
so (2k+m)+(k+m)+(k-m)3=0
so 6k-m=0
so m=6k
so you have a lot of vectors given by
(8i+7j-5k)k for all values of k
(please see that there are 2 k's in the expression i have written .. one is a parameter.. another is the vector along z direction)
24
106That is possible
See bxc is perpendicular to the plane containing b and c.
So its cross product with another vector makes the new vector perpendicular to bxc and hence lie in the same plane of b and c
Similarly ax(bxc) is perpendicular to a
Hence the answer
62any vector perpendicular to bxc will lie in the plane containing both b and c...
also r is perpendicular (orthogonal) to a and (bxc) by the answer..
:)