# If $\pi$ is parallel to $3i-4j$, how to find the vector equation of $\pi$?

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

• $3\hat i - 4\hat j$ is the normal vector. Apr 9 '20 at 3:10
• What relationship must be true between a line in a plane and its normal? Apr 9 '20 at 3:21
• 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. Apr 9 '20 at 3:28
• Apr 9 '20 at 3:34
• @JohnDouma Thanks for the link 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 $$-4x -3y +5z = -18$$

• The answer given is r $\cdot (-4$ i - 3j +5k) = -18 Apr 9 '20 at 3:22
• 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 Apr 9 '20 at 3:33
• $(-i+8j-6k)\cdot(3i-4j)\ne0$, so this answer is incorrect. The second row of the matrix of which you compute the determinant is off.
– amd
Apr 9 '20 at 5:56
• I made a calculation error. Thanks for pointing it out! Apr 9 '20 at 9:09