Cubes, cuboids and planes that cut a segment

What you will be able to do

• place a cube or a cuboid in coordinates and find its diagonals, its centre and its midpoints;
• find the shortest distance from a corner to a diagonal by a quadratic in one variable;
• find the ratio in which a plane parallel to a coordinate plane divides a segment.

A box in a room

A cube is the simplest solid to describe with coordinates. Put one corner at the origin $O$ and the three edges from it along the positive axes. For a cube of side $a$ the eight corners are $(0, 0, 0)$, $(a, 0, 0)$, $(0, a, 0)$, $(0, 0, a)$, $(a, a, 0)$, $(a, 0, a)$, $(0, a, a)$ and $(a, a, a)$. For a cuboid with edges $a$, $b$, $c$, the coordinates are $0$ or $a$, $0$ or $b$, $0$ or $c$.

Now everything is the distance formula.

• Face diagonal of a cube: from $(0, 0, 0)$ to $(a, a, 0)$, of length $a\sqrt{2}$.
• Body diagonal of a cube: from $(0, 0, 0)$ to $(a, a, a)$, of length $a\sqrt{3}$.
• Body diagonal of a cuboid: $\sqrt{a^2 + b^2 + c^2}$. All four body diagonals have this length and meet at the centre.
• Centre: $\left(\dfrac{a}{2}, \dfrac{b}{2}, \dfrac{c}{2}\right)$. It is the midpoint of every body diagonal and is equally far from all eight corners.

Example. A cuboid has edges $4$, $6$ and $12$. Its diagonal is $\sqrt{16 + 36 + 144} = 14$. Its centre is $(2, 3, 6)$, and the distance of the centre from a corner is half the diagonal, $7$.

Midpoints of edges

The midpoint of an edge uses the midpoint formula. For the cube of side $8$, the edge from $(8, 0, 0)$ to $(8, 8, 0)$ has the midpoint $(8, 4, 0)$. The centre is $(4, 4, 4)$, so the distance from the centre to this midpoint is $\sqrt{16 + 0 + 16} = 4\sqrt{2}$.

The shortest distance from a corner to a diagonal

Take the corner $V(a, 0, 0)$ and the body diagonal from $O$ to $(a, a, a)$. A point of the diagonal is $T(ta, ta, ta)$ for $t$ between $0$ and $1$. Then

Complete the square: $3t^2 - 2t + 1 = 3\left(t - \dfrac{1}{3}\right)^2 + \dfrac{2}{3}$. The smallest value is $\dfrac{2}{3}$, at $t = \dfrac{1}{3}$. So the shortest distance is $a\sqrt{\dfrac{2}{3}}$, and the nearest point $T$ is a third of the way along the diagonal. For $a = 6$ the nearest point is $(2, 2, 2)$ and the distance is $2\sqrt{6}$.

This works for any point and any line, by the same method: write the squared distance as a quadratic in $t$ and find its least value.

A plane that cuts a segment

A plane $x = k$ is parallel to the $yz$-plane. It divides the segment from $A(x_1, y_1, z_1)$ to $B(x_2, y_2, z_2)$ in the ratio $(k - x_1) : (x_2 - k)$. The section formula for the $x$-coordinate gives this. If the ratio is positive, the plane cuts inside the segment. The planes $y = k$ and $z = k$ work in the same way.

Example. A drone flies from $A(-2, 1, 4)$ to $B(6, -3, -4)$. It crosses the plane $x = 1$ in the ratio $(1 + 2) : (6 - 1) = 3 : 5$. The crossing point is $\dfrac{3B + 5A}{8} = \left(1, -\dfrac{1}{2}, 1\right)$.

A coordinate plane is the special case $k = 0$, which is the ratio $-x_1 : x_2$ of the section formula lesson.

Practise with new numbers

Try it in Python

The program gives the diagonals of a cube of side $6$. Then it searches many positions of $T$ for the smallest distance $VT$ and compares it with the formula.

Common mistakes

• Using $a\sqrt{2}$ for the body diagonal. The face diagonal is $a\sqrt{2}$. The body diagonal is $a\sqrt{3}$.
• Forgetting the third edge. A diagonal through the inside of a cuboid uses all three edges.
• Swapping the parts of the ratio. For the plane $x = k$ the ratio is $(k - x_1) : (x_2 - k)$, with $A$ first.
• Forgetting to check the sign of the ratio. A negative ratio means that the plane meets the line outside the segment.

Remember these

• Cube of side $a$: face diagonal $a\sqrt{2}$, body diagonal $a\sqrt{3}$, centre $\left(\dfrac{a}{2}, \dfrac{a}{2}, \dfrac{a}{2}\right)$.
• Cuboid: diagonal $\sqrt{a^2 + b^2 + c^2}$.
• The shortest distance from a point to a line: write $d^2$ as a quadratic in $t$ and complete the square.
• Plane $x = k$ cuts $AB$ in the ratio $(k - x_1) : (x_2 - k)$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the face diagonal and the body diagonal of a cube of side $5$.

2. A cuboid has edges $4$, $6$ and $12$ along the axes. Find the length of its diagonal.

3. A cube has side $8$ and one corner at the origin. Find its centre and the distance of the centre from a corner.

4. The plane $z = 3$ cuts the segment from $A(1, 2, -1)$ to $B(5, 6, 7)$. Find the ratio and the point.

5. A cube of side $3$ has a corner at the origin. Find the nearest point of the body diagonal to the corner $(3, 0, 0)$, and the distance.

References

• NCERT, Mathematics Textbook for Class XI, the chapter Introduction to Three Dimensional Geometry, the section on the distance and section formulae.
• OpenStax, Calculus Volume 3, Chapter 2, section 2.2 on vectors in three dimensions.