Given a Vector u= [1 1 2]. Decompose the vector y = [2 3 3] into a sum of two ve

Question

Given a Vector u= [1 1 2]. Decompose the vector y = [2 3 3] into a sum of two vectors, one a multiple of u and the other orthogonal to u.

Explanation / Answer

By decompose, it just means "break into separate vectors that sum to the original vector". ---- You want to find vector projections here. The vector projection of U onto v is given by one of the following identical forms: projv(U) = [ (U•v)/||v|| ] · v/||v|| = [ (U•v)/||v||² ] · v = [ (U•v)/(v•v) ] · v Where ||v|| is the norm (magnitude) of v, and U•v is the dot product of U and v. ---- Obtain the vector projection of A onto B and then subtract that from A to find the perpendicular component. ProjB(A) = [ (A•B)/(B•B) ] · B = [ ( < 1, 0, 3 >•< 5, 2, -6 > )/( < 5, 2, -6 >•< 5, 2, -6 > ) ] · < 5, 2, -6 > = [ ( 1·5 + 0·2 + 3·-6 )/( 5² + 2² + 6² ) ] · < 5, 2, -6 > = [ ( 5 + 0 + -18 )/( 25 + 4 + 36 ) ] · < 5, 2, -6 > = [ ( -13 )/( 65 ) ] · < 5, 2, -6 > = (-1/5)·< 5, 2, -6 > = < -1, -2/5, 6/5 > That is the vector projection, the parallel part. So the perpendicular part is: < 1, 0, 3 > - < -1, -2/5, 6/5 > = < 2, 2/5, 9/5 > So you can express A as vectors parallel and perpendicular to B using: A = projB(A) + [ A - projB(A) ] A = < -1, -2/5, 6/5 > + < 2, 2/5, 9/5 > ---- You can check this result by: (a) making sure that the first vector is a scalar multiple of B (i.e. parallel). Since multiplying the components by -5 gives B, it is parallel. (b) adding the vectors to make sure they equal A.

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