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?

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\$\$

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.

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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