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Let a = (-4, -8, -4) and b = (-3, -2, 0) be vectors. Find the scalar, vector, and orthogonal projections of b onto a. How do I find the orthogonal component?

Also:
Webwork 3 problem 2:
Find a vector orthogonal to both <3, 5, 0> to <0, 5, -2> of the form <1, _, _>
I don't know what finding the normal vector of the normal vector means.

A vector orthogonal to two other vectors will be the cross product between the two vectors.
cross product is defined as:
|--i--j--k--|
|a1a2 a3| = X
|b1b2 b3|

this will be (a2b3-a3b2)i + (a3b1-a1b3)j + (a1b2-b1a2)k = -10i + 6j + 15k = <-10, 6, 15>
multiplying a vector by a constant doesn't change its direction, therefore (-1/10)<-10, 6, 15> is still orthogonal to both.
this vector is <1, -.6, -1.5>


For the first part, the scalar component of b onto a is given by:
(a (dotproduct) b)/|a|
the vector component is the scalar component times a/|a|, the unit vector in the direction of a
the orthogonal component isn't something ive heard referred to like that, but my guess is that it is the vector distance orthogonal to a up to b..
you could find this by doing the scalar component minus b
or by taking a(dotprod)b=|a||b|cos(theta), solving for theta. orthogonal component would be bsin(theta).