find parametric equations for the lines of intersection of the planes: -3x + 2y + z = -5 and 7x + 3y - 2z = -2

Question

find parametric equations for the lines of intersection of the planes: -3x + 2y + z = -5 and 7x + 3y - 2z = -2

Answer 1

u have to put the two of them in order of their variables and then find one variable in terms of the other.. once u have that keep going until u get the answer.. :)

Answer 2

Find parametric equation for the line of intersection of the planes: -3x + 2y + z = -5 and 7x + 3y - 2z = -2

The vector normal to the normal vectors of the planes will be the directional vector v, of the line of intersection of the two planes. Take the cross product.

v = <-3, 2, 1> X <7, 3, -2> = <-7, 1, -23>

Any non-zero multiple of v will also be a directional vector of the line. Multiply by -1.

v = <7, -1, 23>

Now we need to find a point on the line. It will be a point in both planes.

-3x + 2y + z = -5
7x + 3y - 2z = -2

Let x = 0. Now solve for y and z.

2y + z = -5
3y - 2z = -2

Add twice the first equation to the second.

7y = -12
y = -12/7

Plug into the first equation to solve for z.

2(-12/7) + z = -5
-24/7 + z = -5
z = -5 + 24/7 = -11/7

So our point of intersection is P(0, -12/7, -11/7).

So the equation of the line of intersection is:

L(t) = P + tv
L(t) = <0, -12/7, -11/7) + t<7, -1, 23>
where t is a scalar ranging over the real numbers