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The y-axis runs along a straight shore, with all distances in metres…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

The $y$ -axis runs along a straight shore, with all distances in metres. A beacon is placed at a point $P(0, t, 0)$ of the shore. Two buoys float at $A(2, 1, 2)$ and $B(-2, 5, 2)$ . The beacon must be exactly as far from the first buoy as from the second. Find $t$ . Then find the common distance in metres, to two decimal places.

Need a reminder? The key ideaShow

For $PA = PB$ , square both sides. The squares of $x$ , $y$ and $z$ cancel and a plane is left. A fixed ratio $PA = \lambda PB$ ( $\lambda \ne 1$ ) gives a sphere. A point equally far from four points that are not in one plane comes from three linear equations.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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