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Find the shortest distance between the point $Q(11, 2, -1)$ and the line of intersection created by the planes

x .$ \left(\begin{array}{cc} 1\\ -1\\ 3\\ \end{array}\right) = 0 ~$ and x $= a \left(\begin{array}{cc} 2\\ 1\\ 2\\ \end{array}\right)~+ $ $~b \left(\begin{array}{cc} 3\\ 1\\ -3\\ \end{array}\right) $

To get the direction of the line I took $(1, -1, 3)\times(v_1 \times v_2)$ where $v_1$ and $v_2$ are the directional vectors of the second plane.

$ \left(\begin{array}{cc} 1\\ -1\\ 3\\ \end{array}\right) \times$ $ \left(\begin{array}{cc} -5\\ 12\\ -1\\ \end{array}\right) =$ $ \left(\begin{array}{cc} -37\\ -16\\ 7\\ \end{array}\right) $

and the planes go through the origin, so my line is just

$\lambda \left(\begin{array}{cc} -37\\ -16\\ 7\\ \end{array}\right) $

If I imagine a triangle created by the line going from the line of the intersection to the point Q as being the hypotenuse, then if I project Q onto the direction $(-37, -16, 7)$, I get the magnitude of the base of the triangle

$\frac{446}{\sqrt{1674}}$

and all I need to do is use Pythagoras to find the last side of my triangle which will be the perpendicular distance from the line to my point. Therefore I get:

Shortest distance = $\sqrt{126 - (\frac{446}{\sqrt{1674}})^2} \approx 2.678$

I have done it a few times now and keep getting this answer; however, my provided answer is $\sqrt{6} \approx 2.449$. Where am I going wrong?

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  • $\begingroup$ $(1,-1,3) \times (-5,12,-1)$ is $(-35,-14,7)$, not $(-37,-16,7)$. $\endgroup$
    – user856
    Jun 5, 2012 at 4:40

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Your cross product is incorrect. $$\begin{pmatrix} 1 \\ -1 \\ 3\end{pmatrix} \times \begin{pmatrix} -5 \\ 12 \\ -1\end{pmatrix} = \begin{pmatrix} (-1) \times (-1) - 12 \times 3 \\ 3 \times (-5) - 1 \times (-1) \\ 1 \times 12 - (-1) \times (-5)\end{pmatrix} = \begin{pmatrix} -35 \\ -14 \\ 7\end{pmatrix} = 7 \begin{pmatrix} -5 \\ -2 \\ 1\end{pmatrix}$$

As you have correctly observed since both the planes pass through the origin, so does the line of intersection. Hence, if $P$ is the projection of $Q$ on to the line, then $\Delta OPQ$ is a right-angled triangle right-angled at $P$. Hence, the projection of $Q$ onto the line is nothing but $OP$.

Hence, the magnitude of the base of the triangle i.e. $OP$ is $$\dfrac{(11,2,-1) \cdot (-5,-2,1) }{\sqrt{30}} = \dfrac{-55-4-1}{\sqrt{30}} = - \dfrac{60}{\sqrt{30}} = -2 \sqrt{30}$$

Hence, the shortest distance of the point $Q$ to the line is $$PQ = \sqrt{OQ^2 - OP^2} = \sqrt{11^2 + 2^2 +(-1)^2 - \left( -2 \sqrt{30}\right)^2} = \sqrt{126 - 4 \times 30} = \sqrt{6}$$

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