Vectors and Coordinate Geometry: Circles and Spheres

Problem Statement

In standard Cartesian geometry, you learned that the distance $d$ between two points is given by the distance formula, and the midpoint $M$ is the average of their coordinates. You also learned that a circle is defined as the locus of points equidistant from a centre point, leading to the equation $(x-h)^2 + (y-k)^2 = r^2$.

When working in Extension 2 Mathematics, we upgrade these concepts using vectors. The distance between two position vectors $\mathbf{a}$ and $\mathbf{b}$ is simply the magnitude of their difference: $|\mathbf{r} - \mathbf{c}|$. This incredibly simple expression, when set equal to a constant radius $R$, perfectly describes both a circle in 2D space and a sphere in 3D space, vastly simplifying complex geometric problems.

Consider a sphere in 3D space. The points $A$ and $B$ lie on opposite ends of a diameter of the sphere. Their position vectors relative to the origin $O$ are: $\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}$ $\mathbf{b} = 6\mathbf{i} + \mathbf{j} - 3\mathbf{k}$

(a) Using vector methods, find the position vector $\mathbf{c}$ of the centre of the sphere.

(b) Find the radius $R$ of the sphere.

(c) Write down the vector equation of the sphere in the form $|\mathbf{r} - \mathbf{c}| = R$.

(d) Determine if the point $P$ with position vector $\mathbf{p} = 4\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}$ lies inside, outside, or on the surface of the sphere.

Hints

• Part (a): The centre of the sphere is the midpoint of the diameter $AB$. In vector notation, the midpoint of $\mathbf{a}$ and $\mathbf{b}$ is $\frac{1}{2}(\mathbf{a} + \mathbf{b})$.
• Part (b): The radius is half the length of the diameter. Find the vector $\vec{AB} = \mathbf{b} - \mathbf{a}$, calculate its magnitude (the distance formula in 3D), and divide by 2. Alternatively, find the distance from the centre $\mathbf{c}$ to either $\mathbf{a}$ or $\mathbf{b}$.
• Part (c): The vector equation of a sphere is the locus of all position vectors $\mathbf{r}$ such that the distance from $\mathbf{r}$ to the centre $\mathbf{c}$ is exactly $R$. Substitute your answers from (a) and (b).
• Part (d): To test a point, substitute its position vector into the left side of the sphere's equation: $|\mathbf{p} - \mathbf{c}|$. If the result is less than $R$, it's inside; if greater, outside; if equal, on the surface.

Solutions

Part (a): Finding the Centre

1. The centre $\mathbf{c}$ is the midpoint of vectors $\mathbf{a}$ and $\mathbf{b}$.
2. $\mathbf{c} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$ $\mathbf{c} = \frac{1}{2} \left( (2\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}) + (6\mathbf{i} + \mathbf{j} - 3\mathbf{k}) \right)$
3. Add the components: $\mathbf{c} = \frac{1}{2} \left( (2+6)\mathbf{i} + (-3+1)\mathbf{j} + (5-3)\mathbf{k} \right)$ $\mathbf{c} = \frac{1}{2} \left( 8\mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \right)$
4. Divide by 2: $\mathbf{c} = 4\mathbf{i} - \mathbf{j} + \mathbf{k}$

Part (b): Finding the Radius

1. We will find the radius by calculating the distance from the centre $\mathbf{c}$ to the point $A(\mathbf{a})$. $R = |\mathbf{a} - \mathbf{c}|$
2. Calculate the vector difference: $\mathbf{a} - \mathbf{c} = (2\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}) - (4\mathbf{i} - \mathbf{j} + \mathbf{k})$ $\mathbf{a} - \mathbf{c} = (2-4)\mathbf{i} + (-3 - (-1))\mathbf{j} + (5-1)\mathbf{k}$ $\mathbf{a} - \mathbf{c} = -2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}$
3. Find the magnitude of this vector using the 3D distance formula: $R = \sqrt{(-2)^2 + (-2)^2 + (4)^2}$ $R = \sqrt{4 + 4 + 16}$ $R = \sqrt{24} = 2\sqrt{6}$

Part (c): The Vector Equation

1. The general vector equation for a sphere is $|\mathbf{r} - \mathbf{c}| = R$.
2. Substitute the centre vector and the radius: $|\mathbf{r} - (4\mathbf{i} - \mathbf{j} + \mathbf{k})| = 2\sqrt{6}$

Part (d): Testing a Point

1. We need to find the distance from point $P$ to the centre $C$. $\text{Distance} = |\mathbf{p} - \mathbf{c}|$
2. Calculate the vector difference: $\mathbf{p} - \mathbf{c} = (4\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) - (4\mathbf{i} - \mathbf{j} + \mathbf{k})$ $\mathbf{p} - \mathbf{c} = (4-4)\mathbf{i} + (3 - (-1))\mathbf{j} + (2-1)\mathbf{k}$ $\mathbf{p} - \mathbf{c} = 0\mathbf{i} + 4\mathbf{j} + 1\mathbf{k}$
3. Find the magnitude: $|\mathbf{p} - \mathbf{c}| = \sqrt{0^2 + 4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17}$
4. Compare this distance to the radius $R = \sqrt{24}$. Since $\sqrt{17} < \sqrt{24}$, the distance to the point is less than the radius.
5. Therefore, the point $P$ lies inside the sphere.

Takeaways

• Vector Elegance: The expression $|\mathbf{r} - \mathbf{c}| = R$ is vastly cleaner than its Cartesian equivalent $(x-c_x)^2 + (y-c_y)^2 + (z-c_z)^2 = R^2$, yet it holds the exact same geometric meaning.
• Dimensions Don't Matter: The vector midpoint formula $\frac{1}{2}(\mathbf{a}+\mathbf{b})$ and distance formula $|\mathbf{b}-\mathbf{a}|$ work exactly the same way regardless of whether you are working in 2 dimensions, 3 dimensions, or $N$ dimensions.
• Locus Logic: Always read $|\mathbf{r} - \mathbf{c}|$ as "the distance from point $R$ to point $C$". This makes setting up loci equations intuitive.

Further Readings

• HSC Vectors: https://vumaths.com/booklets/hsc-vectors/
• HSC Functions: https://vumaths.com/booklets/hsc-functions/
• HSC Geometry: https://vumaths.com/booklets/hsc-collections/

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Vectors unlock a whole new dimension of geometry. For a deep dive into 3D spaces, check out the comprehensive booklets on Vu's Maths Hub. I post detailed walkthroughs of 3D vector problems on my YouTube channel. For more insights, follow me on Instagram and subscribe to my Substack.