Finding the intersections of 3D planes

**Priori Zenith** · August 22nd 2009, 12:49 AM

Hello I was wondering if you could help with this question:

Find any intersections between the planes:

i) $2x+y-z=0$
and
ii) $z=x^2+y^2+x-3y+15$

Oh and equation (i) and
iii) $(x,y,z) = (3,-5,1)s + (3,0,20)$

Any help at all would be much appreciated. I'm pretty sure you have to sub these equations in to each other, except when I try and do these, I get something like -14=0 and also complex numbers, these are wrong right?

**badgals** · August 22nd 2009, 03:50 AM

i stuck before that. can u help me on how u got to -14=0? but i'm sure if that is the answer then there is no intersections between the planes.

**HallsofIvy** · August 22nd 2009, 03:52 AM

Originally Posted by Priori Zenith

Hello I was wondering if you could help with this question:

Find any intersections between the planes:

i) $2x+y-z=0$
and
ii) $z=x^2+y^2+x-3y+15$

This is not a plane. I presume you mean find the intersection between the plane and the paraboloid. The obvious thing to do is to solve the first equation for z, z= 2x+ y, and set the two equal: $2x+ y= x^2+ y^2+ x- 3y+ 15$ so $x^2+ y^2- x- 4y+ 15= 0$ . Now complete the squares: $x^2- x+ \frac{1}{4}+ y^2- 2y+ 1+ 15- 1- \frac{1}{4}= 0$ and then $(x- \frac{1}{2})^2+ (y- 1)^2= -\frac{55}{4}$ . The sum of squares on the left cannot be negative so this plane and paraboloid do NOT intersect.

Oh and equation (i) and
iii) $(x,y,z) = (3,-5,1)s + (3,0,20)$

Any help at all would be much appreciated. I'm pretty sure you have to sub these equations in to each other, except when I try and do these, I get something like -14=0 and also complex numbers, these are wrong right?

(iii) is a line, not a plane. Yes, you can substitute x= 3s+ 3, y= -5s, z= s+ 20 into the equation of the plane:
2x+y-z= 6s+ 6- 5s- s- 20= 0 which gives -14= 0. That has no solution so, again, there is no point of intersection.

You could also see that by noting that the vector $2\vec{i}+ \vec{j} -\vec{k}$ is perpendicular to the plane 2x+ y- z= 0 while $3\vec{i} -5\vec{j}+ \vec{k}$ points in the direction of the line. The dot product is $2\vec{i}+ \vec{j} -\vec{k}\cdot 3\vec{i} -5\vec{j}+ \vec{k}$ $= 2(3)- 1(5)- 1(1)= 0$ (which is why "s" disappeared from the equation above). Since the line is perpendicular to the vector perpendicular to the plane, the line and plane are parallel and have no intersection.

**Priori Zenith** · August 22nd 2009, 05:41 AM

Thanks so much for that

Just looking though, instead of it being:
$x^2-x+1/4+y^2-2y+1-1/4-1=0$
shouldn't it be:
$x^2-x+1/4+y^2-4y+4-1/4-4=0$

$4y$ instead of $2y$ ??

**badgals** · August 22nd 2009, 05:46 AM

thank you so much

**Priori Zenith** · August 22nd 2009, 05:54 AM

Oh and if I could bother you again,

If I wanted to find the intersection between the line (iii) and the paraboloid (iv) I'd do much the same as I did before, except, when I do this I keep getting a complex number but I'm sure there is an intersection as I have graphed these equations.

**badgals** · August 22nd 2009, 05:59 AM

line(iii) and (iv) does intersect. i've got a quadratic equation for thatas my final answer in terms of s. on the blackboard site, others did confirm that that 2 equations have infinitely many solutions. i did d same for this as with line (i) and (iii), i.e. by substituting the points into (iv). not sure how right is tha though. hope this helps.

**Priori Zenith** · August 22nd 2009, 06:18 AM

Yeah I did what you said, and I got a complex number (i.e. quadratic formula square root was negative), what about the matlab module though, there is an intersection, a complex number means there are no intersections, but there are infinite.

**badgals** · August 22nd 2009, 06:54 PM

erm not sue about that. juz goin to leave the answer as a quadrati equation and say that there are infinitely many solutions

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Finding the intersections of 3D planes

One more thing, sorry.

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