3D geometry question ( Vectors)

aaazureee

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Oct 22, 2015
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The lines l1 and l2 have vector equations
r= (1,-2,3) + t(0,2,1)
r= (1,0,4)+ s(1,-2,1)
respectively for some real parameters t and s

i) Find the acute angle between the two lines l1 and l2. => I have solved this and found the result which is 56.8 degree
ii) Show that l1 passes through point P (1,-4,2)=> Done too!
Hence, show that the distance between point P and any point on the line l2 is given by square root of (6s^2-12s+20). Deduce the shortest distance between point P and line l2.
This part gets confusing since i dont know how to find the general formula of the distance of point P and points on l2? If I can find out the result, then the shortest distance is square root of 14.
I hope someone can help me out :D
 
The lines l1 and l2 have vector equations
r= (1,-2,3) + t(0,2,1)
r= (1,0,4)+ s(1,-2,1)
respectively for some real parameters t and s

i) Find the acute angle between the two lines l1 and l2. => I have solved this and found the result which is 56.8 degree
ii) Show that l1 passes through point P (1,-4,2)=> Done too!
Hence, show that the distance between point P and any point on the line l2 is given by square root of (6s^2-12s+20). Deduce the shortest distance between point P and line l2.
This part gets confusing since i dont know how to find the general formula of the distance of point P and points on l2? If I can find out the result, then the shortest distance is square root of 14.
I hope someone can help me out :D

You found co-ordinate of a point P on line l1.

What would be the co-ordinate of a general point (Q) on l2?
 
The lines l1 and l2 have vector equations
\(\displaystyle \ell_1= (1,-2,3) + t(0,2,1)\)
\(\displaystyle \ell_2= (1,0,4)+ s(1,-2,1)\)
respectively for some real parameters t and s

i) Find the acute angle between the two lines l1 and l2. => I have solved this and found the result which is 56.8 degree
ii) Show that l1 passes through point P (1,-4,2)=> Done too!
Before there can be any angle between two lines there must be a unique point common to both lines. Have you done that?

If that is true then the acute between them is \(\displaystyle \arccos \left( {\dfrac{{\left| {\vec u \cdot \vec v} \right|}}{{\left\| {\vec u} \right\| \cdot \left\| {\vec v} \right\|}}} \right)\) where \(\displaystyle \vec u =<0,2,1>~\&~\vec v=<1,-2,1>\)
 
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