A plane $\pi$ contains the line r = i + $3$j - k +$\lambda(2$i - j + k). If $\pi$ is parallel to $3$i - $4$j, how to find the vector equation of $\pi$?

I know that parallel planes have the same perpendicular vector. But I don’t seem to see how to find any normal vectors to the plane. I can’t cross product two lines, can I? So, how should I solve this question? Thanks

  • $\begingroup$ $3\hat i - 4\hat j$ is the normal vector. $\endgroup$
    – John Douma
    Apr 9 '20 at 3:10
  • $\begingroup$ What relationship must be true between a line in a plane and its normal? $\endgroup$
    – John Douma
    Apr 9 '20 at 3:21
  • $\begingroup$ Actually, I am wrong. I was thinking of two parallel planes that have the same normal vector. A vector parallel to a plane is in the plane so take the cross product. See the answer below. $\endgroup$
    – John Douma
    Apr 9 '20 at 3:28
  • 1
    $\begingroup$ See thejuniverse.org/PUBLIC/LinearAlgebra/LOLA/planes/vect.html $\endgroup$
    – John Douma
    Apr 9 '20 at 3:34
  • $\begingroup$ @JohnDouma Thanks for the link $\endgroup$
    – gc3941d
    Apr 9 '20 at 3:42

Both the vectors $3i - 4j$ and $2i - j + k$ are parallel to the plane, so the normal vector of the plane is perpendicular to both of these vectors. We can find it using the cross product

$$r = \begin{array}{|c c c |} i & j & k \\ 3 & -4 & 0 \\ 2 & -1 & 1\end{array} = -4i-3j+5k$$

This gives us the equation of the plane as $-4x - 3y + 5z= c$ or $4x +3y -5z =d$. The point $i + 3j - k$ lies in this plane, so substituting these values gives us the value of $d$.

$$d = -4 -9 -5 = -18 $$

The equation of the plane is thus $ -4x -3y +5z = -18 $

  • $\begingroup$ The answer given is r $\cdot (-4$ i - 3j +5k) = -18 $\endgroup$
    – gc3941d
    Apr 9 '20 at 3:22
  • $\begingroup$ I think there’s a typo in the first line of the cross product, otherwise your answer would be the same as the book. Thank you $\endgroup$
    – gc3941d
    Apr 9 '20 at 3:33
  • $\begingroup$ $(-i+8j-6k)\cdot(3i-4j)\ne0$, so this answer is incorrect. The second row of the matrix of which you compute the determinant is off. $\endgroup$
    – amd
    Apr 9 '20 at 5:56
  • $\begingroup$ I made a calculation error. Thanks for pointing it out! $\endgroup$ Apr 9 '20 at 9:09

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