1. ## Vector-big question

The straight line L1 with vector equation r=a+tb cuts the plane 2x-3y+6z=6 at the right angles at the point (5,1,-1)

The Line L1= 5i+j-k+t(2i-3j+k). Another straight line, L2, has vector equation r=s(i+3j+2k)

Find the angle between L1 and L2, giving your answer to the nearest degree.

prove that L1 and L2 do not meet, given L2 cuts the plane at the point (-1.2,-3.6,-2.4)

2. Originally Posted by kingkaisai2
The Line L1= 5i+j-k+t(2i-3j+k). Another straight line, L2, has vector equation r=s(i+3j+2k)

Find the angle between L1 and L2, giving your answer to the nearest degree.
Use the following useful fact.
$\displaystyle \bold{u}\cdot \bold{v}=||\bold{u}||\cdot||\bold{v}|| \cos \theta$, where $\displaystyle \theta$ is the angle between these two vectors.

Select any two point on L1, say $\displaystyle t=0,1$
Then you get,
$\displaystyle 5\bold{i}+\bold{j}-\bold{k}=<5,1,-1>$
$\displaystyle 7\bold{i}-2\bold{j}=<7,-2,0>$
The vector between these two points is (in any order),
$\displaystyle \bold{u}=<5-7,1+2,-1-0>=<-2,3,-1>=-2\bold{i}+3\bold{j}-\bold{k}$

For the other line L2, select two point to draw a vector say,
$\displaystyle s=0,1$
Then you get,
$\displaystyle \bold{0}=<0,0,0>$
$\displaystyle \bold{i}+3\bold{j}+2\bold{k}=<1,3,2>$
The vector between these two points is,
$\displaystyle \bold{v}=<1-0,3-0,2-0>=\bold{i}+3\bold{k}+2\bold{k}$

Now do the calculations.
$\displaystyle \bold{u}\cdot \bold{v}=(-2)(1)+(3)(3)+(-1)(2)=5$

$\displaystyle ||\bold{u}||=\sqrt{2^2+3^2+1^2}=\sqrt{14}$
$\displaystyle ||\bold{v}||=\sqrt{1^2+3^2+2^2}=\sqrt{14}$

As a result, you have,
$\displaystyle 5=\sqrt{14}\cdot\sqrt{14}\cos \theta$
Thus,
$\displaystyle 5=14\cos \theta$
Thus,
$\displaystyle \cos\theta =5/14$
Thus,
$\displaystyle \theta\approx 69^o$
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My solution is based on the following observation: the angles between two lines is the same as the angles between any two vectors on these two lines.

3. Originally Posted by ThePerfectHacker
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My solution is based on the following observation: the angles between two lines is the same as the angles between any two vectors on these two lines.
Hello, TPH,

I've got only a question: In my opinion there is only an angle between two lines if they intersect. The given lines don't intersect, they are skew (as I've learned recently of/from/by you), so there doesn't exist any angle and therefore I can't calculate it.

Probably I'm wrong but I can't see where. Asking for help!

Greetings

EB

4. Originally Posted by earboth
Hello, TPH,

I've got only a question: In my opinion there is only an angle between two lines if they intersect. The given lines don't intersect, they are skew (as I've learned recently of/from/by you), so there doesn't exist any angle and therefore I can't calculate it.

Probably I'm wrong but I can't see where. Asking for help!

Greetings

EB
A line can always be represented in the form:

$\displaystyle \bold{l}(s)=\bold{p}+s \bold{u},\ \ \ s \in \mathbb{R}$,

where $\displaystyle \bold{p}$ is a point on the line and $\displaystyle \bold{u}$ is a unit vector along the line.

Now it makes sense to talk about the angle between two lines $\displaystyle \bold{l}_1$
and $\displaystyle \bold{l}_2$ if we mean the angle between the corresponding unit vectors
$\displaystyle \bold{u}_1$ and $\displaystyle \bold{u}_2$.

RonL

5. Originally Posted by earboth
Hello, TPH,

I've got only a question: In my opinion there is only an angle between two lines if they intersect. The given lines don't intersect, they are skew (as I've learned recently of/from/by you), so there doesn't exist any angle and therefore I can't calculate it.

Probably I'm wrong but I can't see where. Asking for help!

Greetings

EB
Here's one possible interpretation. TPH may have another of course.

TPH calculated the direction vectors of the lines and found the angle between those directions. In the picture below, $\displaystyle \bold{b}$ is the direction vector.

Note that when TPH substracted the two points on Line L1= $\displaystyle 5\bold{i}+\bold{j}-\bold{k}+t(2\bold{i}-3\bold{j}+\bold{k})$ he got vector $\displaystyle 2\bold{i}-3\bold{j}+\bold{k}$ with a minus sign. And for Line L2 $\displaystyle \bold{r}=s(\bold{i}+3\bold{j}+2\bold{k})$ he got vector $\displaystyle \bold{i}+3\bold{j}+2\bold{k}$. Those are the direction vectors.

Is the angle between the direction vectors the same as the angle between the lines? Seems right to me. In effect, using the direction vectors translates the lines to a comparable lines through the origin, and then uses the angle between those lines.

6. Originally Posted by JakeD
...

Is the angle between the direction vectors the same as the angle between the lines? Seems right to me. In effect, using the direction vectors translates the lines to a comparable lines through the origin, and then uses the angle between those lines.
Hello, JakeD,

I only aks to be sure that I did understand all: The angle you described is the one which is build by the two direction vectors. This angle will only be visible for a spectator if he looks along the vector $\displaystyle (2,-3,1)\ \times\ (1,3,2)=(-9,-3,9)=k\cdot (-3,-1,3)$. That's the normal vector of the plane which is build by the two direction vectors.
Looking from any other direction you'll see an angle of different size.

Did I get it now correctly? Quite confusing at all!

Greetings

EB

7. Originally Posted by earboth
Hello, JakeD,

I only aks to be sure that I did understand all: The angle you described is the one which is build by the two direction vectors. This angle will only be visible for a spectator if he looks along the vector $\displaystyle (2,-3,1)\ \times\ (1,3,2)=(-9,-3,9)=k\cdot (-3,-1,3)$. That's the normal vector of the plane which is build by the two direction vectors.
Looking from any other direction you'll see an angle of different size.

Did I get it now correctly? Quite confusing at all!

Greetings

EB
Hello, EB.

Yes, that is my understanding too.

Cheers.

JakeD

8. That is an excellent point earboth. I realized that too (in fact the problem says to show they do not intersect) I still assumed that she wanted to calculate it through vectors (as JakeD suggests). For example, two vectors do not have to intersect however they always got an angle betweeen them.