The parametric equations of the line of intersection of the two planes?

Question

The parametric equations of the line of intersection of the two planes
2x+8y+8z=8 and –2x–7y–5z=4
found by Gauss-Jordan elimination are
x= _____+___t
y=______+___t
z=______+___t

Answer 1

Find the line of intersection of the following two planes:
2x + 8y + 8z = 8
-2x - 7y - 5z = 4

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

v = n1 X n2 = <2, 8, 8> X <-2, -7, -5> = <16, -6, 2>

Any non-zero multiple of v will also be a directional vector of the line. Divide by 2.

v = <8, -3, 1>

Now find a point in both planes it will be on the line of intersection. Let z = 0 and solve for x and y.

2x + 8y = 8
-2x - 7y = 4

Adding the two equations we have:

y = 12

Plug back into the first equation and solve for x.

2x + 8y = 8
2x + 8*12 = 8
2x + 96 = 8
2x = 8 - 96 = -88
x = -44

A point P on the line is P(-44, 12, 0).

Now we can write the equation of the line of intersection.

L(t) = P + tv = <-44, 12, 0> + t<8, -3, 1>
L(t) = <-44 + 8t, 12 - 3t, t>
where t is a scalar ranging over the real numbers

Now put the equation of the line in parametric form.

x = -44 + 8t
y = 12 - 3t
z = t

Answer 2

i think u missed one more equation