Find the equation to the plane passing through the points $-2\hat{i}+6\hat{j}-6\hat{k}$, $-3\hat{i}+10\hat{j}-9\hat{k}$ and $-5\hat{i}-6\hat{j}-6\hat{k}$.

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Accepted Answer

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> [!TLDR]
> Two edge vectors give a normal via the cross product; the plane through $A$ with that normal is $12x-3y-8z=6$.
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> **Concept.** A plane is fixed by a point on it and a normal vector. The normal is the cross product of two non-parallel vectors lying in the plane.
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> **Points** $A(-2,6,-6)$, $B(-3,10,-9)$, $C(-5,-6,-6)$.
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> **Edge vectors.**
> $$\overrightarrow{AB}=(-1,4,-3),\qquad \overrightarrow{AC}=(-3,-12,0).$$
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> **Normal $=\overrightarrow{AB}	imes\overrightarrow{AC}$.**
> $$\vec n=\begin{vmatrix}\hat i&\hat j&\hat k\-1&4&-3\-3&-12&0\end{vmatrix}=\hat i(4\cdot0-(-3)(-12))-\hat j((-1)(0)-(-3)(-3))+\hat k((-1)(-12)-4(-3)).$$ …

Q.Find the equation to the plane passing through the points $-2\hat{i}+6\hat{j}-6\hat{k}$ , $-3\hat{i}+10\hat{j}-9\hat{k}$ and $-5\hat{i}-6\hat{j}-6\hat{k}$ .

Concept understanding — Equation of a Plane

Equation of a Plane

Point + normal form

Normal (perpendicular) form

Intercept form …

Q.Find the equation to the plane passing through the points −2i^+6j^−6k^-2\hat{i}+6\hat{j}-6\hat{k}−2i^+6j^​−6k^, −3i^+10j^−9k^-3\hat{i}+10\hat{j}-9\hat{k}−3i^+10j^​−9k^ and −5i^−6j^−6k^-5\hat{i}-6\hat{j}-6\hat{k}−5i^−6j^​−6k^.

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Q.Find the equation to the plane passing through the points $-2\hat{i}+6\hat{j}-6\hat{k}$ , $-3\hat{i}+10\hat{j}-9\hat{k}$ and $-5\hat{i}-6\hat{j}-6\hat{k}$ .

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