Find the equation of the plane parallel to the plane determined by points A, B and C, and passing through the point D





$\overrightarrow{AB} = <1,2,3>$

$\overrightarrow{AC} = <-3,0,0>$

Since $\overrightarrow{AB} \times \overrightarrow{AC} = -9j + 6k$

Is it true that equation of the plane parallel is:




The equation of the plan is given by ${\bf n}\cdot ({\bf x}-{\bf x_0})=0$

Since you found that ${\bf n}=<0,-9,6>$ the equation of the plane become $$<0,-9,6>\cdot (<x,y,z>-(-1,2,4))=-9y+6z-6=0$$

That is $-9y+6z=6$ is the correct answer.

  • 1
    $\begingroup$ Thanks for confirming my answer! I was quite sure I had done something wrong but feels good to know that I didn't. $\endgroup$ – hax0r_n_code Oct 31 '15 at 17:37

the normal is $-9ck+6cj$, $c$ not $0$. Then the equation is $-9c(y-2)+6c(z-4)=0$. Now plug in $D$ to determine $c$.

  • $\begingroup$ I'm sorry but I'm confused by your comments. What is c? $\endgroup$ – hax0r_n_code Oct 31 '15 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.