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**1. Homework Statement**

I know I've had alot of questions about this lately, but this will be the last one...I think I have the right idea but I'm a tad off.

Okay, so the question is

Find a plane containing the point (6,-3,1) and the line of intersection of the planes `-6 x -1 y + 7 z = -10` and `4 x -7 y -6 z = -25`

So you have to find the line, find two points on that line, and then find the plane with the 3 points.

Here are my calculations, but the numbers are funky so I doubt I'm right.

**3. The Attempt at a Solution**

<-6, -1, 7>

<4, -7, -6>

I(6 + 49) - J(36 - 28) + K(42 + 4)

(55, -8, 46)

-1y + 7z = -10

-7y - 6z = -25

-----

-y = -10 - 7z

7(-10 - 7z) - 6z = -25

-70 - 49z - 6z = -25

-55z = 45

z = -(45/55)

-y = -10 - 7(44/55)

-y = -15.72

y = 15.72

z = -.818181

(0, 15.72, -.818181)

r = (0, 15.72, -.818181) + t(55, -8, 46)

or

x = 0 + 55t

y = 15.72 - 8t

z = -.818181 + 46t

-----

Point 1 = (6, -3, 1)

Point 2 = (55, 7.72, 45.181818)

Point 3 = (110, -.28, 91.181818)

A = <51, 10.72, 44.181818>

B = <104, 2.72, 90.181818>

I(966.75 - 120.1745) - J(4599.2727 - 4594.91) + K(138.72 - 1114.88)

I(846.5755) - J(4.3627) + K(-976.16)

846.5755(x - 6) - 4.3627(y + 3) - 976.16(z - 1) = 0

846.5755x - 4.3627y - 976.16z = 5079.453