I'm having some trouble understanding this math problem. It says:
The parallel part is easy. I have that taken care of, but I'm struggling with the orthagonal part. As I understand it, two lines are orthagonal if their dot product is zero. So I found the dot product...
L1 dot L2 = \(\displaystyle \left(4-t\right)\left(\alpha t\right)+\left(-6+2t\right)\left(2-2\alpha t\right)+\left(6+5t\right)\left(3+15t\right)\)
\(\displaystyle -5\alpha t^2+75t^2+16\alpha t+109t+6=0\)
And that's where I'm having difficulty. It seems like the best I can do it express alpha in terms of t. But since t can take on any real value... I don't know how to proceed. The answer key says the lines are orthagonal when \(\displaystyle \alpha=15\). So that leaves me with:
349t + 6 = 0
Which is only true for one value of t, specifically t = -6/349.
51) Find values for \(\displaystyle \alpha\) such that: a) the two lines are parallel; b) the two lines are orthagonal
L1 = \(\displaystyle <4-t,-6+2t,6+5t>\)
L2 = \(\displaystyle <\alpha t,2-2\alpha t,3+15t>\)
for any \(\displaystyle t\in \left(\infty ,-\infty \right)\)
The parallel part is easy. I have that taken care of, but I'm struggling with the orthagonal part. As I understand it, two lines are orthagonal if their dot product is zero. So I found the dot product...
L1 dot L2 = \(\displaystyle \left(4-t\right)\left(\alpha t\right)+\left(-6+2t\right)\left(2-2\alpha t\right)+\left(6+5t\right)\left(3+15t\right)\)
\(\displaystyle -5\alpha t^2+75t^2+16\alpha t+109t+6=0\)
And that's where I'm having difficulty. It seems like the best I can do it express alpha in terms of t. But since t can take on any real value... I don't know how to proceed. The answer key says the lines are orthagonal when \(\displaystyle \alpha=15\). So that leaves me with:
349t + 6 = 0
Which is only true for one value of t, specifically t = -6/349.