Question

(a) Find the orthogonol projection of u on v if u = (3,1,-7) and v = (1,0,5) . (b) Find the vector component of u perpendicular to v . (c) Find IIprojv uII rounded to one decimal place.

Answer 1

i think finding angle between two vectors will do all. \[\cos \theta = u.v \div \left| u \right|\left| v \right|\]
\[v \cos \theta\] would be projection of v on u
\[v \sin \theta\] could be perpendicular component

Is that right?

Answer 2

does that work?

Answer 3

No:(

Answer 4

try to take dot product: (unit vector of u).(v) for projection of v on u

Answer 5

The projection is a little more involved than just a dot product since you want to output a vector. Let us consider vector u and vector v. If we want to find the projection of u on v, p, we are really looking for the following.\[\vec p = \left\lVert \vec u \right\rVert \cos \theta \ \hat v\]Where theta is the angle between the vectors and the unit vector v hat is\[\hat v = \frac{\vec v}{\left\lVert \vec v \right\rVert}\]Let us substitute this in.\[\vec p = \frac{\left\lVert \vec u \right\rVert \cos \theta }{\left\lVert \vec v \right\rVert}\vec v\]The trick is now to multiply the numerator and the denominator by the magnitude of v.\[\vec p = \frac{\left\lVert \vec u \right\rVert\left\lVert \vec v \right\rVert \cos \theta }{\left\lVert \vec v \right\rVert\left\lVert \vec v \right\rVert}\vec v\]Take note that the numerator is now the dot product between u and v and the denominator is the dot product of v with itself.\[\boxed{\displaystyle \vec p = \frac{\vec u \cdot \vec v}{\vec v \cdot \vec v}\vec v}\]This is a formula worthy of memorization. Give it a shot and let me know what you get.