1. ## The Cross Product

Thanks!!

2. Using the normal, n, that you found and the point P
$\displaystyle n = \left[ {\begin{array}{*{20}c} 3 \\ 3 \\ { - 3} \\ \end{array}} \right]\quad P = \left[ {\begin{array}{*{20}c} 1 \\ 0 \\ { - 2} \\ \end{array}} \right]$

write the equation ot the plane
$\displaystyle \Pi :\;n \cdot (R - P) = 0,\quad R = \left[ {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right]$.

3. ## Re:

I'm still a bit unsure...

Equation of the plane for n(3,3,-3)

3x+3y-3x=0

???

4. It should be written as $\displaystyle 3\left( {x - 1} \right) + 3y - 3\left( {z + 3} \right) = 0$.

5. ## Re:

So,

3x-3y-3z=12

Thanks Again,
qbkr21
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}