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 8i7j2k
Then for part b) I used the vector I got for part a) and the point (0,0,0) to get 8x7y2z=0
Both answers are wrong and I don't 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 crossproduct
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 = 2z9
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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