Posted by **Felicia** on Monday, October 1, 2012 at 7:16pm.

Consider the planes given by the equations

2y−2x−z=2

x−2y+3z=7

(a) Find a vector v parallel to the line of intersection of the planes.

(b)Find the equation of a plane through the origin which is perpendicular to the line of intersection of these two planes.

For part a) I just used the cross product of the vectors and got -8i-7j-2k

Then for part b) I used the vector I got for part a) and the point (0,0,0) to get -8x-7y-2z=0

Both answers are wrong and I dont know why.

- Calculus -
**Reiny**, Monday, October 1, 2012 at 7:55pm
Your method is correct, except you must have made an arithmetic error,

I got (-4, -5, -2) as the cross-product

a) which we can change to (4,5,2)

b) since we have a point (0,0,0)

the equation of the required plane is

4x + 5y + 2z = 0

Another way:

Add the two equations

-x + 2z = 9

x = 2z-9

let z = 0, ----> x = -9

sub into 2nd equation, y = -8

let z = 4, ---> x = -1

sub into 2nd equation, y = 2

So we have 2 points on the line of intersection,

(-9, -8, 0) and (-1, 2,4)

and a direction vector would be

(8, 10, 4)

or reduced to (4,5,2) as above

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