Scalar projection of b onto a (vectors)

[TsAmE]

Homework Statement

If a = <3,0,-1> find the vector b such that compaB = 2

Homework Equations

None.

The Attempt at a Solution

$$|a| =\sqrt{3^2 + 1^2} = \sqrt{10}$$

compaB = $$\frac{ a\cdot b}{|a|}$$

$$2 = \frac{3(b1) - 1(b3)}{\sqrt{10}}$$

$$2\sqrt{10} = 3(b1) - 1(b3)$$

I don't know what to do from here

 

[CompuChip]

You said, find the vector b whose projection on a is twice as long as a itself, but of course there are infinitely many such vectors. What you have shown here is that there is that a vector of the form b = (r, s, 3r - 2√10) for any real numbers r and s satisfies comp_ab = 2.

 

[TsAmE]

You said, find the vector b whose projection on a is twice as long as a itself, but of course there are infinitely many such vectors. What you have shown here is that there is that a vector of the form b = (r, s, 3r - 2√10) for any real numbers r and s satisfies comp_ab = 2.

Sorry but I don't understand.