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Consider the following three points in $R^3$ : $P(−1, 1, 0), Q(1, 5, 6), R(3, −1, 4)$

Find the values of $x ∈ R$ for which $PR + x QR$ is perpendicular to $PR$.

I was thinking that equating the dot product of those 2 vectors to $0$ might give the values. But this only gives one x value ($x=-3$) and I presume that there are more. Thoughts?

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Here $\bf{\vec{PR}} = <4,-2,4>$ and $\bf{\vec{QR}} = <2,-6,-2>$ and $\bf{\vec{PR}} = <4,-2,4>$

So $\bf{\vec{PR}+x\vec{QR}} = <4+2x,-2-6x,4-2x>$

Now $\bf{(\vec{PR}+x\vec{QR})\cdot \vec{PR}} =0\Rightarrow (4+2x)\cdot 4+2(2+6x)+4(4-2x) = 0$

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Think you got that spot on.

You can think of the sum $PR + xQR$ as parameterizing a line in $R^3$. The set of guys perpendicular to $PR$ is going to be a plane. Hence you either get one vector (that is, one value of $x$) for which $PR$ and $PR + xQR$ is perpendicular, where the line and the plane intersect, or infinitely many, if the the line is stuck inside the plane.

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