find an equation the the plane. The plane through the point out $(2, −1, 3), (7, 3, 5),$ and also $(−3, −3, −2)$.

You are watching: Find an equation of the plane. the plane through the points (0, 7, 7), (7, 0, 7), and (7, 7, 0)

I acquired the normal vector from $PR * PQ = $and multiply it by the station of the an initial point to acquire $-16x-15y+10z=13$ as response but the answer is wrong. Can someone call me what mistake execute I have?


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There"s one easy way to uncover it.

Solve the determinant,

$$pi : eginvmatrixx -x_1 &y-y_1 &z - z_1 \x_2 -x_1 &y_2-y_1 &z_2 - z_1 \x_3 -x_1 &y_3-y_1 &z_3 - z_1endvmatrix = 0$$

where $x,y,z$ with subscripts denote the works with of the provided three points.

Here us have, $$eginvmatrixx -2&y+1 &z - 3 \5 &4 &2 \-5&-2 &-5endvmatrix = 0$$

$$(x-2)(-16) - (y+1)(-15)+(z-3)(10) = 0$$

$$-16x+15y+10z = -17$$


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When acquisition the cross product, the second term must be reversed, similar to the determinant the a 3x3 matrix.

$$inter-base.netrmPR imes inter-base.netrmPQ= eginvmatrixhatiinter-base.net & hatjinter-base.net & hatk\ 5 &4 &2 \ -5 &-2 &-5 endvmatrix$$

Therefore, it have to be $left langle -16,15,10 ight angle$Then the airplane is $-16x+15y+10z=-17$. We can check that this is undoubtedly the aircraft because every 3 points productivity -17 ~ above the right.


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You space on the right track except one failure in finding your typical vector through your cross product matrix.

The common vector to the airplane is in reality $N= $ instead of $N= $

You have actually forgotten to switch indicators while broadening your matrix.


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