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I have a big problem with geometry. How I do calculate the distance between the vectorial line $r:(x,y,z)=(2,1,0)+\lambda(0,4,-3)$ and the point $A=(2,4,4)$? I tried to solve the problem but nothing...

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  • $\begingroup$ pick a point $P$ on the line so that $AP$ is orthogonal to $(0, 4, -3)$ $\endgroup$ – abel Jan 24 '15 at 16:01
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One way you can calculate it is by taking the distance from $A=(2,4,4)$ to an arbitrary point on the line and then minimizing it.

We let $B_\lambda=(2,1,0)+\lambda(0,4,−3)=(2,1+4\lambda,-3\lambda)$ be an arbitrary point on the line, so using the Euclidean distance formula, we have

$$\begin{align} d(A,B_\lambda)&=\sqrt{(2-2)^2+(4-(1+4\lambda))^2+(4-(-3\lambda))^2} \\&=\sqrt{(3-4\lambda)^2+(4+3\lambda)^2} \\&=\sqrt{9-24\lambda+16\lambda^2+16+24\lambda+9\lambda^2} \\&=\sqrt{25+25\lambda^2} \\&=5\sqrt{1+\lambda^2} \end{align}$$ Since $\lambda^2\ge0$, this is minimized when $\lambda=0$, which gives a distance of exactly $5$.

Note that we have also found the closest point on the line to $A$ which is $B_0=(2,1,0)$.

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  • $\begingroup$ that's awsome, your result is exactly as mine, but the problem is that my professor said that the distance is $3$ not $5$, maybe an error? $\endgroup$ – Ali Mostafa Jan 24 '15 at 16:11
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    $\begingroup$ @AliMostafa Yes, it seems that your professor has made an error. $\endgroup$ – Peter Woolfitt Jan 24 '15 at 16:13

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