Calculus

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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 don't know why.

  • Calculus -

    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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