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Can 3 points be non-coplanar is outside the plane?
If you are given a plane, you can always find and number of points that are not in that plane but, given anythree points there is always at least one plane that goes through all three.
Find the cube root of unity?
The number 1 is its own square root, cube root, etc. in the real number system. To find complex roots of unity, we use the unit circle from trigonometry, coupled with the complex plane, where the x-axis is the real axis, and the y-axis is the imaginary axis. In that coordinate system, the number 1 corresponds to the point (1, 0) and the complex number 1 + 0i. Every complex number a + bi corresponds to the point (a, b) in the complex plane. To find roots of 1, we divide the unit circle up into as many sectors as the number of roots we are trying to find. For cube roots, that's 3 of course, so we divide the unit circle up into 3 sectors of 120 degrees (or 2pi/3 radians) each. So the three cube roots we want are located at the points 120 degrees around the unit circle from (1, 0). Since points on the unit circle have coordinates (cos(theta), sin(theta)), the first one we come to will be (cos(120), sin(120)) = (-1/2, Sqrt(3)/2). This point corresponds to the complex number -1/2 + (sqrt(3)/2)*i. The next point on the circle, 120 degrees from the last one, is (cos(240), sin(240)) = (-1/2, - sqrt(3)/2) = -1/2 - (sqrt(3)/2)*i. Now you have the three cube roots of unity: 1, -1/2 + (sqrt(3)/2)*i, and -1/2 - (sqrt(3)/2)*i. There's much more to all this, involving something called DeMoivre's Formula or Theorem.
What is the multiplicative inverse of 4 i?
The multiplicative inverse of a complex number is found by taking the reciprocal of the number. In this case, the reciprocal of 4i is -1/4i. To find the reciprocal, you divide 1 by the complex number, which results in -1/4i. This is the multiplicative inverse of 4i.
What do you find at the intersection of a plane and a line?
another point
Can you cube root negative two?
yes, you can find a real root to the cube root of any negative real number. There will also be two complex roots which satisfy it, as well.
