Question

the orthogonal projection of the vector u=(-1 -2) on a=(-2 3) is ​

Answer 1

figure: a, b, and the projection of b onto a.

The vector projection of a vector b onto a vector a(figure): we said that the length of the projection is|b| cos(theta), and so, because

|a| |b| cos(theta) = a . b,

we can divide both sides by |a| to get

|b| cos(theta) = the length of the projection = a . b / |a|

The actual vector projection is therefore a unit vector in the correct direction times this length, that is,

projab = (a / |a|)(a . b / |a|).

Next consider the other (unlabelled) vector in the figure. This is the orthogonal projection of b onto a, and its length is (hopefully obviously) |b| sin(theta). Recalling that

|a x b| = |a| |b| sin(theta),

we can find this length by dividing both sides by |a|:

|b| sin(theta) = |a x b| / |a|.

Thus the orthogonal projection orthab =b - projab

Answer 2

$\huge \tt - b$