Question

Linear algebra problem?

Answer 1

Given the versors (i,j,k) in R^3 determine the values of alpha belonging to R so that the vectors u=i+3j+k and v=(alpha+1)i-k form an angle of\[\frac{ \pi }{ 4 }\]

Answer 2

I'd recommend using the dot product formula\[\vec u\cdot\vec v=\|\vec u\|\|\vec v\|\cos\theta_{u,v}\]

Answer 3

I tried using it but in the end the equation doesn't come out right...

Answer 4

If I apply the formula it becomes like this:
\[(\alpha+1)+0-1=\sqrt{2+9}\sqrt{(\alpha+1)^{2}+1}\frac{ \sqrt{2} }{ 2 }\]

Answer 5

After some calculations it becomes like this:
\[2\alpha=\sqrt{11}\sqrt{(\alpha+1)^{2}+1} \sqrt{2}\]

Answer 6

\[4\alpha ^{2}=22((\alpha+1)^{2}+1)\]

Answer 7

\[4\alpha ^{2}=22\alpha ^{2}+44+44\alpha\]

Answer 8

Which then I simplify in\[2\alpha ^{2}-11\alpha ^{2}=22+22\alpha\]

Answer 9

But then I have a problem because the discriminant becomes negative and I can't continue.

Answer 10

Is there something wrong with my calculations?

Answer 11

\[4\alpha ^{2}=22\alpha ^{2}+44+44\alpha\\0=9\alpha^2+22\alpha+22\]the discriminant is then\[22^2-4\cdot9\cdot2>0\]

Answer 12

isn't the discriminant 22^2 - 4 * 9 * [[22]] <-- the [[22]] is the value of c?
i don't think this quadratic has real solutions for a.. and i didn't see anything wrong with the work.

Answer 13

oh wow, I don't guess my hands and brain were not communicating. I guess you guys are right :P

Answer 14

just making some attempt at a graph of the situation. imagining the two vectors, i don't think for any values of alpha they could possibly get narrow enough to make a 45 degree angle (pi/4)

out of sheer curiosity, i also wanted to see a visual of the graph of possible angles just to see where things were looking in comparison to pi/4. poor pi/4...
http://www.wolframalpha.com/input/?i=plot+y+%3D+VectorAngle%5B%7B1%2C3%2C1%7D%2C%7Bx%2B1%2C0%2C-1%7D%5D%2C+y+%3D+pi%2F4

Answer 15

Sorry if I reply right now but OpenStaudy didn't work for me yesterday.
So is my work correct or not?

Answer 16

I agree with all your work. It seems that there is indeed no values for alpha unless there is some typo not accounted for.

Answer 17

So at the point you realized that the discriminant of the quadratic was negative and no real solutions existed for alpha, that would be the point where we can say that no values of alpha work.

Answer 18

Ah, ok. I hope it's like that then. My book doesn't have solutions of the problems so I can't check if I'm doing things right or wrong. thanks in advance.

Answer 19

Err...I meant thank you. I'm falling asleep xD