Kendra F. answered • 08/26/16
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The general equation of a plane is:
n • < x - x0, y - y0, z - z0 > = 0
One point (2,0,3) is provided. You'll need 3 points total.
n • < x - x0, y - y0, z - z0 > = 0
One point (2,0,3) is provided. You'll need 3 points total.
Set t = 0
(-1,0,-4)
Set t = 1
(0,1,-2)
Set t = 1
(0,1,-2)
Let "a" be the vector from (0,1,-2 ) to (-1,0,-4); a = < -1, -1, -2 > (found by subtracting coordinates)
Let "b" be the vector from (0,1,-2 ) to (2,0,3); b = < 2, -1, 5 >
The cross product, a x b = n, normal vector
< -1, -1, -2 > x < 2, -1, 5 > = < -7, 1, 3 >
Use the general equation of a plane, the point (2,0,3) and the normal vector to find the Eq of the plane.
n • < x - x0, y - y0, z - z0 > = 0
n • < x - x0, y - y0, z - z0 > = 0
< -7, 1, 3 > • < x - 2, y - 0, z - 3 >
-7(x - 2) + 1(y - 0) + 3(z - 3) = 0
-7x + 14 + y - 0 + 3z - 9 = 0
-7x + y + 3z = -5
Kendra F.
The point (-1,0,4) does not fall on the line. At t = 0 the point is (-1, 0, -4)
This is why your answer is off.
This is why your answer is off.
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08/26/16
Pia R.
oh right so you mean it doesnt matter what i choose like something like a reference point?
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08/26/16
Kendra F.
Yea. It shouldn't matter because all the points lay on the plane. I reworked the problem using the same points as you (with the added negative sign) and arrived at the same answer then noticed the missing sign. Make sure to check your steps carefully and good luck!
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08/26/16
Pia R.
08/26/16