Find parametric equations of the line perpendicular to the line through the three points P=(0,1,0), Q=(-1,1,2) and R=(2,1,-1) and intersecting the plane at the point P=(3,1,-2).

Thank you

2. Originally Posted by HorseStar2
Find parametric equations of the line perpendicular to the line through the three points P=(0,1,0), Q=(-1,1,2) and R=(2,1,-1) and intersecting the plane at the point P=(3,1,-2).

Thank you
use formula for line that goes through 3 points

$\begin{vmatrix}
x-x_1&y-y_1 &z-z_1 \\
x_2-x_1&y_2-y_1 &z_2-z_1 \\
x_3-x_1&y_3-y_1 &z_3-z_1
\end{vmatrix}$

and to intersect plane at any (let's say your point) line that you get from that up there must have point (P(3,1,-2))

hint: draw it for your self ... it make more sense that way

P.S. actually as i more think of it... this could be some "problem" if there is equation of the plane so you need to find where that line intersects (if intersect) plane ... but this way when you solve that determinant up there there is not much to do ... if you have 3 points there is no another way to draw line through them except like that, and there is no another condition by which you can lean it so it will go through plane at that point ...

funny thing is that you can substitute any from P,Q,R with that P on the plane and put in determinant and you will get the same result

3. Originally Posted by HorseStar2
Find parametric equations of the line perpendicular to the line through the three points P=(0,1,0), Q=(-1,1,2) and R=(2,1,-1) and intersecting the plane at the point P=(3,1,-2).

Thank you
Are you sure you have the right P, Q and R. I believe that they're not colinear.

4. Originally Posted by Danny
Are you sure you have the right P, Q and R. I believe that they're not colinear.
Hi, I made an error when writing the question, it should read:

Find parametric equations of the line perpendicular to the PLANE through the three points P=(0,1,0), Q=(-1,1,2) and R=(2,1,-1) and intersecting the plane at the point P=(3,1,-2).

Do I have to find the parametric equations of the two lines PQ and PR, then calculate the cross product of the two directional vectors to give a common normal to the plane? I'm not sure where to go from here if that's right....

Thanks

5. Originally Posted by yeKciM
use formula for line that goes through 3 points

$\begin{vmatrix}
x-x_1&y-y_1 &z-z_1 \\
x_2-x_1&y_2-y_1 &z_2-z_1 \\
x_3-x_1&y_3-y_1 &z_3-z_1
\end{vmatrix}$

and to intersect plane at any (let's say your point) line that you get from that up there must have point (P(3,1,-2))

hint: draw it for your self ... it make more sense that way

P.S. actually as i more think of it... this could be some "problem" if there is equation of the plane so you need to find where that line intersects (if intersect) plane ... but this way when you solve that determinant up there there is not much to do ... if you have 3 points there is no another way to draw line through them except like that, and there is no another condition by which you can lean it so it will go through plane at that point ...

funny thing is that you can substitute any from P,Q,R with that P on the plane and put in determinant and you will get the same result
Hi, I made an error when writing the question, it should read:

Find parametric equations of the line perpendicular to the PLANE through the three points P=(0,1,0), Q=(-1,1,2) and R=(2,1,-1) and intersecting the plane at the point P=(3,1,-2).

Do I have to find the parametric equations of the two lines PQ and PR, then calculate the cross product of the two directional vectors to give a common normal to the plane? I'm not sure where to go from here if that's right....

Thanks

6. sorry if i miss translate it to my self that "perpendicular to the PLANE" i think that means that line is normal to the plane ?
if so your equation of that line is :
$y=1$

or am I of with this, and it's plane trough 3 points

7. Originally Posted by HorseStar2
Hi, I made an error when writing the question, it should read:

Find parametric equations of the line perpendicular to the PLANE through the three points P=(0,1,0), Q=(-1,1,2) and R=(2,1,-1) and intersecting the plane at the point P=(3,1,-2).

Do I have to find the parametric equations of the two lines PQ and PR, then calculate the cross product of the two directional vectors to give a common normal to the plane? I'm not sure where to go from here if that's right....

Thanks
You're on the right track! Find the vector $\vec{PQ}$ and $\vec{PR}.$ Then cross these (like you said). This will give the normal to the plane. The line perpendicular to the plane will follow this normal. Then use the point (3,1,-2) to finally give you your line.

8. Originally Posted by Danny
You're on the right track! Find the vector $\vec{PQ}$ and $\vec{PR}.$ Then cross these (like you said). This will give the normal to the plane. The line perpendicular to the plane will follow this normal. Then use the point (3,1,-2) to finally give you your line.
Hi,

So far I have calculated.. vector PQ = (-1, 0, 2) and Vector PR = (2, 0, -1) so corresponding parametric lines are: PQ = (0, 1, 0 ) + t1(-1, 0, 2), PR = (0, 1, 0) + t2 (2, 0, -1). Completing the cross product of the vectors PQ and PR I obtained 0i + 3j + 0k. Is this right so far? so do I then have a final answer of (3, 1, -2) + t3(0, 3, 0) as my parametric equation of the line?

Thank you

9. Originally Posted by HorseStar2
Hi,

So far I have calculated.. vector PQ = (-1, 0, 2) and Vector PR = (2, 0, -1) so corresponding parametric lines are: PQ = (0, 1, 0 ) + t1(-1, 0, 2), PR = (0, 1, 0) + t2 (2, 0, -1). Completing the cross product of the vectors PQ and PR I obtained 0i + 3j + 0k. Is this right so far? so do I then have a final answer of (3, 1, -2) + t3(0, 3, 0) as my parametric equation of the line?

Thank you
When I write (3, 1, -2) + t3(0, 3, 0) ... t3 is t subscript 3

10. Yep - that's it!

11. Originally Posted by Danny
Yep - that's it!
Really... you're not just saying that lol?