Question

Let O be the origin and OA =a1i + a2j + a3k. Find the equation of the planes such that every point on the plane is equidistant from the two end of the vector OA. You are asked to do this by first finding the point of intersection of the plane with the vector OA. State the condition for vector OA to have a magnitude iof 25

Answer 1

Answer on Question #63630 – Math – Analytic Geometry

Question

Let $O$ be the origin and $\overrightarrow{OA} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k}$. Find the equation of the planes such that every point on the plane is equidistant from the two ends of the vector $\overrightarrow{OA}$. You are asked to do this by first finding the point of intersection of the plane with the vector $\overrightarrow{OA}$. State the condition for vector $\overrightarrow{OA}$ to have a magnitude of 25.

Solution

Let $M(x,y,z)$ be a point in $\mathbb{R}^3$. Then

$$OM^2 = x^2 + y^2 + z^2;$$ $$AM^2 = (x - a_1)^2 + (y - a_2)^2 + (z - a_3)^2.$$

Point M is equidistant from the two ends of the vector $\overrightarrow{OA}$.

$$x^2 + y^2 + z^2 = (x - a_1)^2 + (y - a_2)^2 + (z - a_3)^2;$$ $$x^2 + y^2 + z^2 = x^2 + y^2 + z^2 - 2a_1x - 2a_2y - 2a_3z + a_1^2 + a_2^2 + a_3^2;$$ $$a_1x + a_2y + a_3z - \frac{a_1^2 + a_2^2 + a_3^2}{2} = 0.$$

The result is a plane perpendicular to the vector $\overrightarrow{OA} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k}$.

The midpoint N of the vector $\overrightarrow{OA}$ is the point of intersection of the plane with the vector $\overrightarrow{OA}$.

Indeed, $N\left(\frac{a_1}{2}, \frac{a_2}{2}, \frac{a_3}{2}\right)$:

$$\frac{a_1^2}{2} + \frac{a_2^2}{2} + \frac{a_3^2}{2} - \frac{a_1^2 + a_2^2 + a_3^2}{2} \equiv 0.$$

If the vector $\overrightarrow{OA}$ has a magnitude of 25, then

$$a_1^2 + a_2^2 + a_3^2 = 25^2;$$ $$a_1^2 + a_2^2 + a_3^2 = 625.$$

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