Find the direction cosines of the a vector in the plane of vec(b) =<2,1,3> and vec(c)=<−1,2,1> and perpendicular to vec(c)

charlygyloavao9 2022-10-08 Answered
Find the direction cosines of the a vector in the plane of b =< 2 , 1 , 3 > and c =< 1 , 2 , 1 > and perpendicular to c
Let the vector a = x , y , z
Since it is perpendicular to c
x + 2 y + z = 0
And
| x y z 2 1 3 1 2 1 |
5 x 7 y + 5 z = 0
since a, b, c are coplanar
So y=0
Then x=y
So the answer should be 1 2 , 0 , 1 2 , but that is not correct.
What is wrong with my solution?
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Answers (2)

Piper Pruitt
Answered 2022-10-09 Author has 9 answers
Let r = p b + q c , take dot product of this with c , and use r . c = 0, then
r . c = p b . c + q c . c p = q c 2 b . c = q 6 3 = 2 q .
So the required vector r is:
r = q [ 2 b + c ] = q ( 5 i 5 k ) .
q is arbitrary real number. Next r ^ = i + k 2
It also means that
r = α   c × ( b × c ) ,
namely, the vector triple product.
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Aidyn Crosby
Answered 2022-10-10 Author has 3 answers
Note that the projection of b onto the direction perpendicular to c is
b ( b c ) c | c | 2 =< 2 , 1 , 3 > 1 2 < 1 , 2 , 1 >=< 5 2 , 0 , 5 2 >
Normalize to get a =< 1 2 , 0 , 1 2 >
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