Angle between line l and line m is 60degree

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SUMMARY

The problem involves finding the value of 'a' such that the angle between line l, defined by the equation r = s(1,2,-1), and line m, defined by r = (0,1,1) + t(5,a,5), is 60 degrees. The solution utilizes the formula for the cosine of the angle between two vectors, leading to the equation √6√(50 + a²)cos(60) = 2a. After correcting an initial error in the vector cross product, the final answer for 'a' is determined to be √30.

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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) = [tex]\sqrt{1^2 + 2^2 + 1^2}[/tex]x[tex]\sqrt{5^2 + a^2 + 5^2}[/tex] x cos 60
 
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Yes, [itex]\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)[/itex].
So [itex]\sqrt{6}\sqrt{50+ a^2}cos(60)= 10+ 2a[/itex]
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, [itex]\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)[/itex].
So [itex]\left(\sqrt{6}\sqrt{50+ a^2}cos(60)= 10+ 2a[/itex]
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 = [tex]\sqrt{30}[/tex]. 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.
 

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