Angle between line l and line m is 60degree

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Homework Statement


The line l has the equation r= s(1,2,-1) and the line m has the equation r= (0,1,1)+t(5,a,5). Find the value of a such that the angle between the line l and m is 60degree.

The Attempt at a Solution


What I did was use the equation s(1,2,-1)xt(5,a,-5) = \sqrt{1^2 + 2^2 + 1^2}x\sqrt{5^2 + a^2 + 5^2} x cos 60
 
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Yes, \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta).
So \sqrt{6}\sqrt{50+ a^2}cos(60)= 10+ 2a
All you have to do is solve that equation for a. I recommend squaring both sides to get rid of the square root and solving the resulting quadratic equation.
 
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HallsofIvy said:
Yes, \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta).
So \left(\sqrt{6}\sqrt{50+ a^2}cos(60)= 10+ 2a
All you have to do is solve that equation for a. I recommend squaring both sides to get rid of the square root and solving the resulting quadratic equation.
Yea finally I am able to solve it. I was having dilemma on the square root part but your advice had been really helpful. Thanks a lot HallsofIvy. Cheers :smile:
I got my answer a = \sqrt{30}. I accidentally put s(1,2,-1)xt(5,a,-5) when its suppose to be s(1,2,-1)xt(5,a,5) therefore its suppose to be 2a instead of 10+2a.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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