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Find straight line L' perpendicular to L, contained in plane $\Pi$ and that has (-1, 6, 1).

(-1, 6, 1) is intersection between $L$ and $\Pi$

$\Pi$ is $X + Y - 2Z = 3$

$L$ is $t(-2, 2, 2) + (0, 5, 0)$

I was trying to do this, but I don't know how to put 3 conditions in the equation.

On the other hand, for L' to be contained in $\Pi$, then L' must be perpendicular to (1, 1, -2), which is Normal Vector of $\Pi$.

But then L is not perpendicular to $\Pi$ (because (1, 1, -2)*(-2, 2, 2) is not 0) so how is it possible that L' is perpendicular to L and be contained in $\Pi$?

Is it a typo in the exam? Because later I have to find intersection between L and L'

1 Answers 1

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The direction of $L$ is $(-2,2,2)$

And the normal of $\Pi$ is $(1,1,-2)$

You need a vector that is perpendicular to both of those.

Any idea?

Try the cross product.

The intersection of $L$ and $L'.$ $L'$ lies in the plane $\Pi$ and L intersect the plane $\Pi$ at only one point.

Now you have a vector, and you have the point $(-1,6,1)$ on the line, can you generate the equation of the line?

  • 1
    I have no problem with that, but will the vector be contained in $\Pi$ ? – 2017-01-31
  • 2
    If the vector is perpendicular to the normal of $\Pi$ it must, at the very least, be parallel to $\pi$. And if it shares a point in $\Pi$ then it must lie in $\Pi$ – 2017-01-31