Calculus III Vector Problems

[riverjib]

Hello,

Can somebody help me with the following two problems?

1. Find a value of a such that the vector <a, 1,1> makes an angle of 45 degrees with the vector <1,2,1> or show that no such a exists.

---I started by using theta = arccos (u*v)/(absu*absv) must equal (sqrt2)/2 and attempting to solve from there...but I'm not sure if that's the best way to proceed.

2. Let u = 3i + j; v = 5i - 2j; and w = i - j. Find scalars, a and b such that u = av +bw.

I have no idea how to proceed with this one.

I would greatly appreciate any help!!!

 

[o_O]

1. It's not theta that equals to \(\displaystyle \frac{\sqrt{2}}{2}\) but:
\(\displaystyle cos\theta = \frac{\sqrt{2}}{2}\)

You never had to deal with the arccos function in the first place :wink: So just use your dot product formula and solve for a:
\(\displaystyle \vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cdot \frac{\sqrt{2}}{2}\)
\(\displaystyle (a)(1) + (1)(2) + (1)(1) = \sqrt{a^{2} + 1^{2} + 1^{2}} \cdot \sqrt{1^{2} + 2^{2} + 1^{2}} \cdot \frac{\sqrt{2}}{2}\)
etc. etc.

2. Well, you get two systems of equations:
\(\displaystyle \left(\begin{array}{c}3 \\ 1 \end{array} \right) = a\left(\begin{array}{c}5 \\ -2 \end{array} \right) + b\left(\begin{array}{c}1 \\ -1 \end{array} \right)\)

\(\displaystyle 3 = 5a + b\)
\(\displaystyle 1 = -2a - b\)
You can either add the two equations or solve for one variable in terms of the other. Your pick!

 

[riverjib]

Thanks!!