In Specialist Mathematics Unit 4, vectors are extended from two-dimensional space into three-dimensional space (\(\mathbb{R}^3\)). This involves defining lines and planes using vector notation, which provides a powerful alternative to Cartesian coordinate geometry.
A line in three-dimensional space is uniquely determined by a point on the line and a direction vector parallel to the line.
The vector equation of a line passing through a point with position vector \(\mathbf{a}\) and parallel to a direction vector \(\mathbf{d}\) is given by:
\$\(\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}, \quad \lambda \in \mathbb{R}\)\$
Where:
* \(\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) is the position vector of any general point on the line.
* \(\mathbf{a} = x_1\mathbf{i} + y_1\mathbf{j} + z_1\mathbf{k}\) is a known point on the line.
* \(\mathbf{d} = l\mathbf{i} + m\mathbf{j} + n\mathbf{k}\) is the direction vector.
* \(\lambda\) is a scalar parameter.
By equating the \(\mathbf{i, j, k}\) components of the vector equation, we obtain the parametric equations:
\$\(x = x_1 + \lambda l\)\$
\$\(y = y_1 + \lambda m\)\$
\$\(z = z_1 + \lambda n\)\$
By solving each parametric equation for \(\lambda\) and equating them, we derive the Cartesian form:
\$\(\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}\)\$
Note: If one of the direction components is zero (e.g., \(l=0\)), the line lies in a plane parallel to a coordinate plane (e.g., \(x = x_1\)).
EXAM TIP: When finding the equation of a line passing through two points \(A\) and \(B\), remember that the direction vector \(\mathbf{d}\) is simply the displacement vector \(\vec{AB} = \mathbf{b} - \mathbf{a}\).
A plane in 3D can be defined in several ways, most commonly by a point in the plane and a vector perpendicular to the surface.
If a plane contains a point with position vector \(\mathbf{a}\) and is perpendicular to a normal vector \(\mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), any point \(\mathbf{r}\) on the plane satisfies:
\$\((\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0\)\$
Rearranging gives the standard vector equation of a plane:
\$\(\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}\)\$
Let \(\mathbf{a} \cdot \mathbf{n} = d\) (a scalar constant). The equation becomes:
\$\(\mathbf{r} \cdot \mathbf{n} = d\)\$
Expanding the scalar product \(\mathbf{r} \cdot \mathbf{n} = d\) where \(\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) and \(\mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) results in:
\$\(ax + by + cz = d\)\$
The coefficients \((a, b, c)\) are the components of the normal vector to the plane.
A plane can also be defined by a point \(\mathbf{a}\) and two non-parallel (linearly independent) vectors \(\mathbf{u}\) and \(\mathbf{v}\) that lie within the plane:
\$\(\mathbf{r} = \mathbf{a} + \lambda \mathbf{u} + \mu \mathbf{v}, \quad \lambda, \mu \in \mathbb{R}\)\$
KEY TAKEAWAY: The normal vector \(\mathbf{n}\) is the most critical piece of information for a plane. If you are given two vectors \(\mathbf{u}\) and \(\mathbf{v}\) in the plane, you can find the normal vector using the cross product: \(\mathbf{n} = \mathbf{u} \times \mathbf{v}\).
To find where the line \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}\) intersects the plane \(\mathbf{r} \cdot \mathbf{n} = d\):
1. Substitute the expression for \(\mathbf{r}\) from the line equation into the plane equation.
2. Solve for the scalar \(\lambda\): \((\mathbf{a} + \lambda \mathbf{d}) \cdot \mathbf{n} = d\).
3. \(\mathbf{a} \cdot \mathbf{n} + \lambda(\mathbf{d} \cdot \mathbf{n}) = d \implies \lambda = \frac{d - \mathbf{a} \cdot \mathbf{n}}{\mathbf{d} \cdot \mathbf{n}}\).
4. Substitute \(\lambda\) back into the line equation to find the position vector of the intersection point.
| Relationship | Condition |
|---|---|
| Line parallel to Plane | Direction vector \(\mathbf{d}\) is perpendicular to normal \(\mathbf{n}\): \(\mathbf{d} \cdot \mathbf{n} = 0\) |
| Line perpendicular to Plane | Direction vector \(\mathbf{d}\) is parallel to normal \(\mathbf{n}\): \(\mathbf{d} = k\mathbf{n}\) |
| Two Planes are Parallel | Normal vectors are parallel: \(\mathbf{n_1} = k\mathbf{n_2}\) |
| Two Planes are Perpendicular | Normal vectors are perpendicular: \(\mathbf{n_1} \cdot \mathbf{n_2} = 0\) |
COMMON MISTAKE: Students often assume that if a line is parallel to a plane, its direction vector is parallel to the normal. In fact, if a line is parallel to a plane, it is perpendicular to the plane’s normal vector.
All angle calculations in 3D rely on the scalar product formula: \(\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\).
The angle \(\theta\) between two lines is the angle between their direction vectors \(\mathbf{d_1}\) and \(\mathbf{d_2}\):
\$\(\cos(\theta) = \frac{|\mathbf{d_1} \cdot \mathbf{d_2}|}{|\mathbf{d_1}||\mathbf{d_2}|}\)\$
The angle \(\theta\) between two planes is the angle between their normal vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}\):
\$\(\cos(\theta) = \frac{|\mathbf{n_1} \cdot \mathbf{n_2}|}{|\mathbf{n_1}||\mathbf{n_2}|}\)\$
The angle \(\alpha\) between a line with direction \(\mathbf{d}\) and a plane with normal \(\mathbf{n}\) is the complement of the angle between \(\mathbf{d}\) and \(\mathbf{n}\). If \(\theta\) is the angle between \(\mathbf{d}\) and \(\mathbf{n}\):
\$\(\sin(\alpha) = \cos(\theta) = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}\)\$
VCAA FOCUS: VCAA frequently asks for the “acute” angle. Always use the absolute value of the dot product in the numerator to ensure \(\cos(\theta) > 0\), resulting in an acute angle.
The shortest (perpendicular) distance from a point \(P\) with position vector \(\mathbf{p}\) to the plane \(\mathbf{r} \cdot \mathbf{n} = d\) is:
\$\(\text{Distance} = \frac{|\mathbf{p} \cdot \mathbf{n} - d|}{|\mathbf{n}|}\)\$
If the plane is in Cartesian form \(ax + by + cz = d\) and the point is \((x_0, y_0, z_0)\):
\$\(\text{Distance} = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}\)\$
Substituting \(\mathbf{p} = \mathbf{0}\) into the formula:
\$\(\text{Distance} = \frac{|d|}{|\mathbf{n}|}\)\$
REMEMBER: The unit normal vector \(\hat{\mathbf{n}} = \frac{\mathbf{n}}{|\mathbf{n}|}\) is useful here. The equation \(\mathbf{r} \cdot \hat{\mathbf{n}} = p\) represents a plane where \(p\) is the perpendicular distance from the origin to the plane.
