1. ## angle between lines

given the two lines calculate the angle that is formed by the intersection of the pair of lines

L1:x=-1+3t,y=1+4t,z=-2t
L2:x=-1+2s,y=3s,z=-7+s

i have tried everything even converting these to symmetric equation and figuring out the Cartesian coordiantes for both lines which are for L1(-1,1,0) and for L2(-1,0,7)

then of course taking the dot product and dividing by the magnitude of both and the negative cosine to get the angle

2. Originally Posted by william
given the two lines
L1:x=-1+3t,y=1+4t,z=-2t
L2:x=-1+2s,y=3s,z=-7+s
Find the angle between $<3,4,-2>~\&~<2,3,1>$

3. Originally Posted by Plato
These two are skew lines.
So what do you mean by "an angle between them"?
what are skew lines?

the question states calculate the angle that is formed by the intersection of each pair of lines

pardon me for not clearly stating the question, my mistake

4. Originally Posted by william
what are skew lines?

the question states calculate the angle that is formed by the intersection of each pair of lines

pardon me for not clearly stating the question, my mistake
Sorry I made a mistake. They intersect.
Why don’t you bother to learn basic LaTeX??
http://www.mathhelpforum.com/math-he...-tutorial.html

5. Originally Posted by Plato
Find the angle between $<3,4,-2>~\&~<2,3,1>$
i get $\frac{16}{\sqrt 29 \sqrt 14}=\frac{16}{20.14}=0.79=37.4^o$ after taking the dot product and dividing by both lines magnitudes

also i have learned basic LaTex, i apologize for my laziness

also, thanks for the help

also, can you verify my work?

6. $\arccos \left( {\frac{{16}}
{{\sqrt {29} \sqrt {14} }}} \right) = 0.653 = 37.433^ \circ
$

7. Originally Posted by Plato
$\arccos \left( {\frac{{16}}
{{\sqrt {29} \sqrt {14} }}} \right) = 0.653 = 37.433^ \circ
$
thanks a lot for your help, perhaps you can have a look at my very similar problem that would be must appreciated

http://www.mathhelpforum.com/math-he...een-lines.html

8. Hello, william!

Given the two lines, calculate the angle formed by the intersection of the lines.

. . $L_1\!:\;\;\begin{Bmatrix}x&=&-1+3t \\ y&=&1+4t \\ z&=& -2t\end{Bmatrix}\qquad L_2\!:\;\begin{Bmatrix}x&=&-1+2s \\ y&=&3s \\z&=&-7+s \end{Bmatrix}$
$\text{Formula: }\;\cos\theta \;=\;\frac{\vec u \cdot \vec v}{|\vec u|\,|\vec v|}$

We have: . $\begin{array}{ccc}\overrightarrow{L_1} &=& \langle 3,4,\text{-}2\rangle \\ \\[-4mm] \overrightarrow{L_2} &=& \langle 2,3,1\rangle \end{array}$

Hence: . $\cos\theta \;=\;\frac{\langle3,4,-2\rangle \cdot\langle 2,3,1\rangle}{\sqrt{9+16+4}\,\sqrt{4+9+1}} \;=\;\frac{6+12-2}{\sqrt{29}\,\sqrt{14}} \;=\;\frac{16}{\sqrt{406}} \;=\;0.794066667$

Therefore: . $\theta \;=\;37.43281131^o \;\approx\;37.4^o$