Intersection of line and plane

[apple53]

Intersection of line and plane!

Homework Statement

Intersection of line and plane!
Okay i to find the common points of line and plane

Question r=i+j+A(2k-j) and r . (i+j) = 4

Homework Equations

I heard that it is easier to use the vector equation in the form r . n = p

The Attempt at a Solution

So if i do that to above

You get

1 0 1

1 + A -1 . 1 =4

0 2 0

The above doesn't sound right i won't get an x, y, z value i defo need help on this or can anyone start my off

kind regards

And also how do you wrap the vectors in brackets on this forum lol

 

[HallsofIvy]

Homework Statement

Intersection of line and plane!
Okay i to find the common points of line and plane

Question r=i+j+A(2k-j) and r . (i+j) = 4

Homework Equations

I heard that it is easier to use the vector equation in the form r . n = p

The Attempt at a Solution

So if i do that to above

You get

1 0 1

1 + A -1 . 1 =4

0 2 0

It's not clear to me what the first and third lines are but it looks like the second line is almost the result of putting r=i+j+A(2k-j)= i+ (1- A)jk+ 2Ak into the equation r . (i+j) = 4.
Rather than what you have, you should get 1+ (1- A)= 2- A= 4.

Solve That for A, then put that value into the equation of the line.

The above doesn't sound right i won't get an x, y, z value i defo need help on this or can anyone start my off

kind regards

And also how do you wrap the vectors in brackets on this forum lol

 

[apple53]

Sorry about the way its laid out. your right that's what i have done

i put r=i+j+A(2k-j) into r . (i+j) = 4

which gives me

i+j+A(2k-j) . (i+j) = 4

I tried writing this in vectorial format but don't khow how to get the brackets to wrap i,j,k on this website. If you can help with that i would be gratefull

Anyway to continue on with question we break them down in components. Correct me if I am wrong

1+A(0) . 1=4
1+A(-1) . 1=4
0+A(2) . 0=4

Which is where i got previously. Which is probably wrong

Rather than what you have, you should get 1+ (1- A)= 2- A= 4.

Now to continue with what you did I am trying to work out how you derived the above.

I broke the vector r=i+j+A(2k-j) into its components ready to put the A values in

so you get

1+A(0)=X This gives x=1
1+A(-1)=Y
0+A(2)=Z

Just need to know how you derived A whcih i quotes above. If you did put put r=i+j+A(2k-j) into r . (i+j) = 4 To get the equation i+j+A(2k-j) . (i+j) = 4. And from this you managed to get A. Can you please explain to me step by step please

Thanks for quick reply and Kinds regards