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

Question

Accepted Answer

>

> [!TLDR]
> Take two edge vectors from the points, cross them for a normal $(-12,3,8)$, and write the plane through $A$: $12x-3y-8z=6$.
>
> **Concept.** A plane through three points has normal $\vec n=\vec{AB}	imes\vec{AC}$; the equation is $\vec n\cdot(\vec r-\vec A)=0$.
>
> The position vectors give points $A(-2,6,-6)$, $B(-3,10,-9)$, $C(-5,-6,-6)$.
>
> **Step 1 — edge vectors.**
> $$\vec{AB}=(-1,4,-3),\qquad \vec{AC}=(-3,-12,0).$$
>
> **Step 2 — normal (cross product).**
> $$\vec n=\vec{AB}	imes\vec{AC}=\begin{vmatrix}\hat i&\hat j&\hat k\-1&4&-3\-3&-12&0\end{vmatrix}=\hat i(0-36)-\hat j(0-9)+\hat k(12+12)=(-36,\,9,\,24).$$ …

Q.Find the equation of a 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 of a 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^.

🎓Unlock everything free for 14 days

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

Unlock everything free for 14 days