I assume you mean the length of the projection. Recall that the projection of u onto v is u•v/|v|. So, since the line y=x is in the direction of i+j+0k, we have
|2i-5j+k • (i+j)/√2| = 3/√2
Find projection of 2i-5j+k on line y=x
4 answers
recall that a
projection of vector a on b = ( (a dot b)/|b|^2 )(vector b)
let a = <2, -5, 1>
b = <1, 1, 1> ..... (the equation says y = x, which I did with <1,1,...> and since there is no mention of z I can let it be anything )
a dot b = 2-5+1 = -3
and |b|^2 = ( √(1+1+1) )^2 = 3
so the projection of 2i-5j+k on line y=x = (-3/3)(vector b)
= (-1)<1,1,1> = <-1,-1,-1>
projection of vector a on b = ( (a dot b)/|b|^2 )(vector b)
let a = <2, -5, 1>
b = <1, 1, 1> ..... (the equation says y = x, which I did with <1,1,...> and since there is no mention of z I can let it be anything )
a dot b = 2-5+1 = -3
and |b|^2 = ( √(1+1+1) )^2 = 3
so the projection of 2i-5j+k on line y=x = (-3/3)(vector b)
= (-1)<1,1,1> = <-1,-1,-1>
I think we both messed up. The projection of A onto B is
A•b where b is the unit vector in the direction of B: B/|B|
A•B/|B|
since A•B = |A|*|B| cosθ
A•b where b is the unit vector in the direction of B: B/|B|
A•B/|B|
since A•B = |A|*|B| cosθ
My error was in using a vector which satisfies y = x
I should have used <1,1,0> like oobleck did
The projection of vector a on b = ( (a dot b)/|b|^2 )(vector b)
the component of that projection vector is (a dot b)/|b|
So the projection vector is (<2, -5, 1> dot <1, 1, 0> )/√2 * <1,1,0>
= -3/√2 <1,1,0>
I should have used <1,1,0> like oobleck did
The projection of vector a on b = ( (a dot b)/|b|^2 )(vector b)
the component of that projection vector is (a dot b)/|b|
So the projection vector is (<2, -5, 1> dot <1, 1, 0> )/√2 * <1,1,0>
= -3/√2 <1,1,0>