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enter image description here

Problem

In the picture above are given the coordinates of the points $O(0,0,0)$, $A(6,0,0)$, $C(0,12,0)$, $D(0,0,5)$, $K(0,6,5)$, $L(6,12,4)$, $M(6,8,0)$, $N(0,8,0)$.

It seems as if line $KL$ is parallel to plane $DEM$. Show if they are indeed parallel or not.

My attack

I set up an equation for plane $DEM$: $5y+8z=40$.

Then, I set up a vector equation for line $KL$: $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\6\\5\end{pmatrix}+\lambda\begin{pmatrix}6\\6\\-1\end{pmatrix}$.

I plugged in $y=6+6\lambda$ and $z=5-\lambda$ into the equation for plane $DEM$, and ended up with $\lambda=-\frac{15}{11}$.

This means they intersect at a point with $\lambda=-\frac{15}{11}$.

But, my book says they are parallel. Am I wrong or is my book wrong?

Thanks for the help.

@Minibill, how do you explain the visible intersection in this plot of the line and the plane?

enter image description here

rae306
  • 9,742

2 Answers2

2

You are quite correct. Following your line of reasoning, if $\lambda=-\frac{15}{11},$ then we have $$\begin{pmatrix}x\\y\\z\end{pmatrix}= \begin{pmatrix}-90/11\\-24/11\\70/11\end{pmatrix},$$ which can be shown to lie in the given plane.

It is possible that there is a typo in the problem, or that you are reading the solution to a different problem, but it appears you have done everything correctly.

Cameron Buie
  • 102,994
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You are right: an intersection just needs to satisfy the plane equation and the laws you stated for y and z (it must also satisfy the one for x, that is $x = 6 \lambda$, but here it has no influence because, as you observed, the equation for the plane does not contain $x$).

EDIT: first answer was completely wrong because I misread an equation

miniBill
  • 278