Plane Equation Finding 2
Find the equation of the plane containing the points $ (0,-1,-1), $ $ (-4,4,4), $ and $ (4,5,1) $. Enter your answer in the form
\[Ax + By + Cz + D = 0,\]where $ A, $ $ B, $ $ C, $ $ D $ are integers such that $ A > 0 $ and $ \gcd(|A|,|B|,|C|,|D|) = 1 $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Let $ \mathbf{a} = \begin{pmatrix} 0 \\ -1 \\ -1 \end{pmatrix}, $ $ \mathbf{b} = \begin{pmatrix} -4 \\ 4 \\ 4 \end{pmatrix}, $ and $ \mathbf{c} = \begin{pmatrix} 4 \\ 5 \\ 1 \end{pmatrix} $. Then the normal vector of the plane is orthogonal to both
\[\mathbf{b} - \mathbf{a} = \begin{pmatrix} -4 \\ 5 \\ 5 \end{pmatrix}\]and
\[\mathbf{c} - \mathbf{a} = \begin{pmatrix} 4 \\ 6 \\ 2 \end{pmatrix}.\]So to compute the normal vector, we take the cross product of these vectors:
\[\begin{pmatrix} -4 \\ 5 \\ 5 \end{pmatrix} \times \begin{pmatrix} 4 \\ 6 \\ 2 \end{pmatrix} = \begin{pmatrix} -20 \\ 28 \\ -44 \end{pmatrix}.\]We can scale this vector, and take $ \begin{pmatrix} 5 \\ -7 \\ 11 \end{pmatrix} $ as the normal vector. Then the equation of the plane is of the form
\[5x - 7y + 11z + D = 0.\]Substituting the coordinates of any of the points, we find that the equation of the plane is $ \boxed{5x - 7y + 11z + 4 = 0} $.
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