Distances, angles and intersections
Master distances, angles and intersections for Year 12 Specialist Mathematics in Queensland (QCAA). Using the dot and cross products, you measure the shortest distance from a point to a plane or line, the angle between lines, planes and a line and a plane, and the point where a line pierces a plane.
You will learn to read the normal and direction vectors off any equation, choose the matching formula, and work each distance and angle in full — the geometric payoff of 3D vectors and a core skill for the rest of Unit 3.
Theory
Distances, angles and intersections are the measuring tools of 3D vector geometry in Year 12 Specialist Mathematics (QCAA, Queensland, Unit 3). Using the dot product and the cross product, you find the shortest distance from a point to a plane or line, the angle between two lines, two planes or a line and a plane, and where lines and planes intersect. This page collects every formula and shows each one worked in full.
Once a line is written \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}\) and a plane as \(\mathbf{r}\cdot\mathbf{n}=d\) (equivalently \(ax+by+cz=d\)), the normal \(\mathbf{n}\) and the direction \(\mathbf{d}\) carry all the geometry. Distances and angles then reduce to dot and cross products.
Distances are always measured along the perpendicular — the shortest route. From a point to a plane use \(\dfrac{|\mathbf{n}\cdot\mathbf{p}-d|}{|\mathbf{n}|}\); from a point to a line use \(\dfrac{|\overrightarrow{AP}\times\mathbf{d}|}{|\mathbf{d}|}\); between two skew lines use \(\dfrac{|(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2)|}{|\mathbf{d}_1\times\mathbf{d}_2|}\).
Angles come from the same vectors. Two lines meet at the angle between their directions; two planes meet at the angle between their normals; a line and a plane use \(\sin\theta=\dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}\), because the line-plane angle is the complement of the angle to the normal. An absolute value keeps every angle acute.
Intersections are solved algebraically. A line meets a plane where its parametric coordinates satisfy the plane equation — substitute and solve for \(\lambda\). Two planes meet in a line whose direction is \(\mathbf{n}_1\times\mathbf{n}_2\), perpendicular to both normals.
For a plane \(\mathbf{r}\cdot\mathbf{n}=d\) and a point \(P\) with position \(\mathbf{p}\), the perpendicular distance is:
For a line through \(A\) with direction \(\mathbf{d}\), the distance from a point \(P\) is:
The three angle formulas — two lines, two planes (via normals), and a line with a plane:
Two parallel planes \(\mathbf{r}\cdot\mathbf{n}=d_1\) and \(=d_2\) are apart by:
Choose and apply the right formula
- Read off the vectors. Take the normal \(\mathbf{n}\) from the plane's \(x,y,z\) coefficients and the direction \(\mathbf{d}\) from the line's parameter; note the fixed point \(\mathbf{a}\) of each line.
- Name the measurement. Decide whether you need a distance (point-plane, point-line, skew, parallel planes), an angle (line-line, plane-plane, line-plane), or an intersection point.
- Substitute into the matching formula. Compute the dot or cross product, then its magnitude; for an intersection, substitute the line's coordinates into the plane and solve for \(\lambda\).
- Take the absolute value and simplify. Keep distances positive and angles acute with \(|\cdots|\); rationalise surds and give exact values unless a degree answer is asked.
Read the normal off the coefficients and find its length:
| \(\mathbf{n}\) | \(=\) | \((2,1,-2)\) |
| \(|\mathbf{n}|\) | \(=\) | \(\sqrt{4+1+4}=3\) |
Compute \(\mathbf{n}\cdot\mathbf{p}\), then apply \(\dfrac{|\mathbf{n}\cdot\mathbf{p}-d|}{|\mathbf{n}|}\):
| \(\mathbf{n}\cdot\mathbf{p}\) | \(=\) | \((2)(4)+(1)(2)+(-2)(3)\) |
| \(=\) | \(8+2-6=4\) | |
| \(\text{dist}\) | \(=\) | \(\dfrac{|4-1|}{3}\) |
| \(=\) | \(\dfrac{3}{3}=1\) |
The distance from \(P\) to the plane is \(1\).
Take the base point \(A(1,0,1)\) and form \(\overrightarrow{AP}\):
| \(\overrightarrow{AP}\) | \(=\) | \(P-A\) |
| \(=\) | \((2-1,\ 4-0,\ 1-1)\) | |
| \(=\) | \((1,4,0)\) |
Cross \(\overrightarrow{AP}\) with the direction \(\mathbf{d}=(2,1,2)\):
| \(\overrightarrow{AP}\times\mathbf{d}\) | \(=\) | \((4\cdot 2-0\cdot 1,\ 0\cdot 2-1\cdot 2,\ 1\cdot 1-4\cdot 2)\) |
| \(=\) | \((8,-2,-7)\) |
Apply \(\dfrac{|\overrightarrow{AP}\times\mathbf{d}|}{|\mathbf{d}|}\):
| \(|\overrightarrow{AP}\times\mathbf{d}|\) | \(=\) | \(\sqrt{64+4+49}=\sqrt{117}\) |
| \(=\) | \(3\sqrt{13}\) | |
| \(|\mathbf{d}|\) | \(=\) | \(\sqrt{4+1+4}=3\) |
| \(\text{dist}\) | \(=\) | \(\dfrac{3\sqrt{13}}{3}=\sqrt{13}\) |
The distance from \(P\) to the line is \(\sqrt{13}\).
Write the line's coordinates in terms of \(t\):
| \(x\) | \(=\) | \(2+t\) |
| \(y\) | \(=\) | \(t\) |
| \(z\) | \(=\) | \(1+2t\) |
Substitute into the plane equation and solve for \(t\):
| \((2+t)-(t)+(1+2t)\) | \(=\) | \(5\) |
| \(3+2t\) | \(=\) | \(5\) |
| \(2t\) | \(=\) | \(2\) |
| \(t\) | \(=\) | \(1\) |
Put \(t=1\) back into the line:
| \((x,y,z)\) | \(=\) | \((2+1,\ 1,\ 1+2)\) |
| \(=\) | \((3,1,3)\) |
The line meets the plane at \((3,1,3)\).
Read the normal, then use \(\sin\theta=\dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}\):
| \(\mathbf{n}\) | \(=\) | \((2,2,1)\) |
| \(\mathbf{d}\cdot\mathbf{n}\) | \(=\) | \((2)(2)+(-1)(2)+(2)(1)\) |
| \(=\) | \(4-2+2=4\) |
Divide by the two magnitudes and invert:
| \(|\mathbf{d}|\) | \(=\) | \(3,\quad |\mathbf{n}|=3\) |
| \(\sin\theta\) | \(=\) | \(\dfrac{4}{3\times 3}=\dfrac{4}{9}\) |
| \(\theta\) | \(=\) | \(\sin^{-1}\!\left(\dfrac{4}{9}\right)\) |
| \(\approx\) | \(26^\circ\) |
The support makes about \(26^\circ\) with the plane.
Common pitfalls
Frequently asked questions
How do you find the distance from a point to a plane?
Write the plane as \(\mathbf{r}\cdot\mathbf{n}=d\), then use \(\dfrac{|\mathbf{n}\cdot\mathbf{p}-d|}{|\mathbf{n}|}\), where \(\mathbf{p}\) is the point's position vector. The absolute value keeps the distance positive, and dividing by \(|\mathbf{n}|\) measures along the perpendicular.
How do you find the distance from a point to a line in 3D?
Pick any point \(A\) on the line, form \(\overrightarrow{AP}\), and use \(\dfrac{|\overrightarrow{AP}\times\mathbf{d}|}{|\mathbf{d}|}\). The cross product's magnitude is the area of the parallelogram, and dividing by \(|\mathbf{d}|\) leaves the perpendicular height.
Why is the line-plane angle formula sine and not cosine?
The formula \(\dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}\) gives the angle between the line and the plane's normal. The line-plane angle is the complement of that, and the complement swaps cosine for sine, so \(\sin\theta=\dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}\).
How do you find where a line meets a plane?
Write the line's coordinates \(x,y,z\) in terms of the parameter, substitute them into the plane equation, and solve for the parameter. Putting that value back into the line gives the point of intersection.
How do you find the angle between two planes?
Use their normals: \(\cos\theta=\dfrac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}\). The absolute value gives the acute angle between the planes. If the normals are perpendicular the planes are perpendicular.
What is the shortest distance between two skew lines?
It is measured along the common perpendicular: \(\dfrac{|(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2)|}{|\mathbf{d}_1\times\mathbf{d}_2|}\). The cross product \(\mathbf{d}_1\times\mathbf{d}_2\) points along that perpendicular direction.