Find the vector projection of u onto v. Then write u as the sum of two orthogonal​ vectors, one of which is projvu. u=〈−2​,3〉​, v=〈5​,6〉

Question

Find the vector projection of u onto v. Then write u as the sum of two orthogonal​ vectors, one of which is proj_vu. u=〈−2​,3〉​, v=〈5​,6〉

Answer 1

The projection of vector u onto vector v, denoted as proj_v u, is given by the formula:

proj_v u = ((u . v) / ||v||^2) * v

where:

1. "." denotes the dot product of vectors u and v,
2. "||v||" is the magnitude of vector v.

First, calculate the dot product of vectors u and v:

u . v = (-25) + (36) = -10 + 18 = 8

Next, find the magnitude of vector v:

||v|| = sqrt((5^2) + (6^2)) = sqrt(25 + 36) = sqrt(61)

Now, find the square of the magnitude of vector v:

||v||^2 = 61

Now, find the projection of u onto v:

proj_v u = ((u . v) / ||v||^2) * v = (8 / 61) * 〈5, 6〉 = 〈(8/61)*5, (8/61)*6〉 = 〈0.656, 0.787〉

Now, we can write vector u as the sum of two orthogonal vectors, one of which is proj_v u.

The other vector is the difference between vector u and the projection, proj_v u, and it's orthogonal to v. Let's call this vector w:

w = u - proj_v u = 〈-2, 3〉 - 〈0.656, 0.787〉 = 〈-2.656, 2.213〉

So, the vector u can be written as the sum of two orthogonal vectors:

u = proj_v u + w = 〈0.656, 0.787〉 + 〈-2.656, 2.213〉 = 〈-2, 3〉

This confirms that vector u can be broken down into two vectors, one of which is the projection of u onto v (and thus parallel to v) and the other is orthogonal to v.