HELP Intersection of two lines (VECTORS)

[apple53]

HELP! Intersection of two lines (VECTORS)

Homework Statement

Find the common point of the lines r=i+j+k+x(j-3k) and r=i+y(k-j)

Homework Equations

The Attempt at a Solution

If the lines intersect then there are numbers x and y such that

i+j+k+x(j-3k)=i+y(k-j)

The two lines above does intersect at a point. But my question is how did they find that point!. I am guessing you must need to find x and y to find the point BUT HOW!

kinds regards

I will send the right answer after somebody attempts the above question which i need help on

 

[DiracRules]

I think it could be helpful to see a way to solve if you put the lines in vectorial notation:
$r: \left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\1\\1\end{array}\right]+x\cdot\left[\begin{array}{c}0\\1\\-3\end{array}\right]$
$s: \left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\0\\0\end{array}\right]+y\cdot\left[\begin{array}{c}0\\-1\\1\end{array}\right]$

Now, it should be easy to find the intersection...

(Just check if I wrote correctly the vectors)

 

[apple53]

I can see what you mean. I wrote the vectorial notation below. Since we are writing it in this way we will have to change the x and y before parallel vector into other letter e.g. a and b. So i wrote the Vectorial notation as it should be below

r:⎡⎣xyz⎤⎦=⎡⎣111⎤⎦+a⋅⎡⎣01−3⎤⎦
r:⎡⎣xyz⎤⎦=⎡⎣100⎤⎦+b⋅⎡⎣0−11⎤⎦
This does help

From here i would write

x=1+a(0) (from this we get x=1)
x=1+b(0)

y=1+a(1)=1+a
y=0+b(-1)=-b

z=1+a(-3)=1-3a
z=0+b(1)=b

We need to find a or b to help find y and z. I have chosen to find a. I did simultaneous equations for both y to get 1-3a=-1-a. Which rearranges to give 1+1=3a-a which gives a=1

I put a in above equation to get y. e.g. y=1+1(1)=2

and put a in above equation to get z e.g. z=1-3(1)=-2

Which give me the coordinates (1,2,-2)

which is the correct answer

Thanks DiracRules