Calc III: Two Intersecting Planes?

Question

Find the equation of the plane through point (-1, 4, 2) that contains the line of intersection of the planes:
4x - y + z - 2 = 0 and
2x + y - 2z - 3 = 0

By making y = t, I found the intersecting line equation to be:
x = 7/10 + 1/10 t
y = t
z = -4/5 + 3/5 t.

So this direction is <1/10, 1, 3/5>

Where do I go from here? The answer in the back of the book is:
4x - 13y + 21z + 14 = 0

I'd appreciate any help.

Thank you very much dust, but how did you know to write the 2 given planes in parametric form
4x-y+z-2+t(2x+y-2z-3) = 0? My book doesn't discuss how to do problems of this type. I follow everything after that though.

Answer 1

Find the equation of the plane through point P(-1, 4, 2) that contains the line of intersection of the planes:
4x - y + z - 2 = 0 and
2x + y - 2z - 3 = 0
_________

This amounts to finding the plane that contains a given point and a given line. They just haven't told you what the line is. You need to figure that out. So that's where we will start.

The directional vector v, of the line of intersection is normal to the normal vectors n1 and n2, of the two planes. Take the cross product.

v = n1 X n2 = <4, -1, 1> X <2, 1, -2> = <1, 10, 6>

Now we just need to find a point on the line. For the equations of the two planes, let x = 0 and solve for y and z.

-y + z - 2 = 0
y - 2z - 3 = 0

Add the two equations.

-z - 5 = 0
-z = 5
z = -5

Plug the value of z into the first equation and solve for y.

-y - 5 - 2 = 0
-y = 7
y = -7

So a point on the line of intersection is Q(0, -7, -5).

The line of intersection is:

L(t) = P + tv = <0, -7, -5> + t<1, 10, 6>

Now create a second directional vector u, for the plane.

u = PQ = = <0+1, -7-4, -5-2> = <1, -11, -7>
u = <1, -11, -7>

The normal vector n, of the desired plane is normal to both of the directional vectors of the plane. Take the cross product.

n = u X v = <1, -11, -7> X <1, 10, 6> = <4, -13, 21>

With the normal vector of the plane and a point on the plane we can write the equation of the plane. Let's choose
P(-1, 4, 2).

4(x + 1) - 13(y - 4) + 21(z - 2) = 0
4x + 4 - 13y + 52 + 21z - 42 = 0
4x - 31y + 21z + 14 = 0

Answer 2

3 Intersecting Planes

Answer 3

ok this is how it is done
find the general equation of the plane passing through the line of intersection of the two planes ...the equation would be
4x-y+z-2+t(2x+y-2z-3)=0
simplifying
(4+2t)x+(t-1)y+(1-2t)z+(-2-3t)=0------------(1)
now this passes through (-1,4,2)...substituting in (1)
we get t= -8/5

(4/5)x+(-13/5)y+(21/5)z+14/5=0
thus the equation of the plane is
4x-13y+21z+14=0

Answer 4

f(x,y) = x² + y² is a cylinder with z axis as its axis. Pl y = xtanT is // to z axis passing thr the commencing place. So the sections would be 2 a million/2 cylinders. think of that a cylindrical cake is cut back alongside the diameter of its actual around face.