Vectors in Three Dimensions: Your Exam Success Guide

What the examiner is testing

The examiner is assessing your ability to translate geometric relationships in three-dimensional space into vector equations and then manipulate these equations to find specific points, distances, or angles. Marks are typically awarded for correctly setting up vector equations for lines and planes, performing scalar products to determine perpendicularity or angles, and solving simultaneous equations to find intersection points.

The method

1. Identify the type of vector: Determine if the question refers to position vectors (from the origin to a point) or displacement vectors (between two points). Represent position vectors using \(\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) or as a column vector \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\).
2. Formulate vector equations for lines: A line passing through point A (with position vector \(\mathbf{a}\)) and parallel to vector \(\mathbf{b}\) has the equation \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\), where \(\lambda\) is a scalar parameter.
3. Formulate vector equations for planes (if required): A plane can be defined by a point on the plane and a normal vector, or by three non-collinear points. The equation of a plane with normal \(\mathbf{n}\) passing through point A (position vector \(\mathbf{a}\)) is \(\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}\).
4. Calculate the magnitude of a vector: For \(\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\), the magnitude is \(|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}\). This is often used for distances or unit vectors.
5. Compute the scalar (dot) product: For two vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\), the scalar product is \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\). This is crucial for finding angles and checking perpendicularity (\(\mathbf{a} \cdot \mathbf{b} = 0\)).
6. Determine the angle between two vectors: Use the formula \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\), where \(\theta\) is the angle between vectors \(\mathbf{a}\) and \(\mathbf{b}\).
7. Find the intersection of lines or a line and a plane: Equate the vector equations for the coordinates (\(x, y, z\)) and solve the resulting simultaneous equations for the parameters (\(\lambda, \mu\), etc.). Substitute the parameter values back into one of the original equations to find the intersection point.

Worked example

A line \(L\) passes through the point \(A(1, 2, 3)\) and is parallel to the vector \(2\mathbf{i} - \mathbf{j} + 4\mathbf{k}\). Find the position vector of the point \(P\) on \(L\) such that \(P\) is 3 units away from \(A\).

First, write the equation of the line \(L\):
$$ \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} $$
Let the position vector of point \(P\) be \(\mathbf{p}\). Then \(\vec{AP}\) is given by:
$$ \vec{AP} = \mathbf{p} - \mathbf{a} = \lambda \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} $$
We are given that the distance \(AP\) is 3 units, which means \(|\vec{AP}| = 3\).
$$ \left| \lambda \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} \right| = 3 $$
$$ |\lambda| \left| \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} \right| = 3 $$
Calculate the magnitude of the direction vector:
$$ \left| \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} \right| = \sqrt{2^2 + (-1)^2 + 4^2} = \sqrt{4 + 1 + 16} = \sqrt{21} $$
So,
$$ |\lambda| \sqrt{21} = 3 $$
$$ |\lambda| = \frac{3}{\sqrt{21}} $$
This gives two possible values for \(\lambda\): \(\lambda = \frac{3}{\sqrt{21}}\) or \(\lambda = -\frac{3}{\sqrt{21}}\).
Let's use \(\lambda = \frac{3}{\sqrt{21}}\) to find one possible position vector for \(P\):
$$ \mathbf{p} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \frac{3}{\sqrt{21}} \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} $$
$$ \mathbf{p} = \begin{pmatrix} 1 + \frac{6}{\sqrt{21}} \\ 2 - \frac{3}{\sqrt{21}} \\ 3 + \frac{12}{\sqrt{21}} \end{pmatrix} $$
Rationalising the denominators:
$$ \mathbf{p} = \begin{pmatrix} 1 + \frac{6\sqrt{21}}{21} \\ 2 - \frac{3\sqrt{21}}{21} \\ 3 + \frac{12\sqrt{21}}{21} \end{pmatrix} = \begin{pmatrix} 1 + \frac{2\sqrt{21}}{7} \\ 2 - \frac{\sqrt{21}}{7} \\ 3 + \frac{4\sqrt{21}}{7} \end{pmatrix} $$
The position vector of \(P\) is \(\left(1 + \frac{2\sqrt{21}}{7}\right)\mathbf{i} + \left(2 - \frac{\sqrt{21}}{7}\right)\mathbf{j} + \left(3 + \frac{4\sqrt{21}}{7}\right)\mathbf{k}\).
(The other solution would be for \(\lambda = -\frac{3}{\sqrt{21}}\), giving a point on the other side of A).

Sanity check: The distance \(AP\) should be 3.
\(\vec{AP} = \frac{3}{\sqrt{21}} \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\).
\(|\vec{AP}| = \left| \frac{3}{\sqrt{21}} \right| \left| \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} \right| = \frac{3}{\sqrt{21}} \times \sqrt{21} = 3\). This matches the given information.

Worked example: a harder one

The points \(A\), \(B\) and \(C\) have position vectors \(\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\), \(\mathbf{b} = 4\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) and \(\mathbf{c} = \mathbf{i} + 5\mathbf{j} - 2\mathbf{k}\) respectively.
Find the shortest distance from the point \(C\) to the line \(L\) passing through \(A\) and \(B\).

Why the obvious first move fails: A common first thought is to find the vector \(\vec{AC}\) and then calculate its magnitude. This gives the distance from \(A\) to \(C\), but not the shortest distance from \(C\) to the line \(L\). The shortest distance is the perpendicular distance.

Correct approach:
1. Find the vector equation of the line \(L\):
The line passes through \(A\) and \(B\). The direction vector \(\vec{AB}\) is \(\mathbf{b} - \mathbf{a}\).
$$ \vec{AB} = (4\mathbf{i} + 2\mathbf{j} + \mathbf{k}) - (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) = 2\mathbf{i} + 3\mathbf{j} - 2\mathbf{k} $$
The equation of line \(L\) is \(\mathbf{r} = \mathbf{a} + \lambda \vec{AB}\):
$$ \mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 3 \\ -2 \end{pmatrix} $$
2. Define a general point \(P\) on the line \(L\):
Let \(P\) be a point on \(L\) with position vector \(\mathbf{p}\).
$$ \mathbf{p} = \begin{pmatrix} 2 + 2\lambda \\ -1 + 3\lambda \\ 3 - 2\lambda \end{pmatrix} $$
3. Form the vector \(\vec{CP}\):
$$ \vec{CP} = \mathbf{p} - \mathbf{c} = \begin{pmatrix} 2 + 2\lambda \\ -1 + 3\lambda \\ 3 - 2\lambda \end{pmatrix} - \begin{pmatrix} 1 \\ 5 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 + 2\lambda \\ -6 + 3\lambda \\ 5 - 2\lambda \end{pmatrix} $$
4. Use the condition for shortest distance: The shortest distance from \(C\) to \(L\) occurs when \(\vec{CP}\) is perpendicular to the direction vector of \(L\), which is \(\vec{AB}\). This means their scalar product is zero.
$$ \vec{CP} \cdot \vec{AB} = 0 $$
$$ \begin{pmatrix} 1 + 2\lambda \\ -6 + 3\lambda \\ 5 - 2\lambda \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 3 \\ -2 \end{pmatrix} = 0 $$
$$ 2(1 + 2\lambda) + 3(-6 + 3\lambda) - 2(5 - 2\lambda) = 0 $$
$$ 2 + 4\lambda - 18 + 9\lambda - 10 + 4\lambda = 0 $$
$$ 17\lambda - 26 = 0 $$
$$ \lambda = \frac{26}{17} $$
5. Substitute \(\lambda\) back into \(\vec{CP}\) to find the perpendicular vector:
$$ \vec{CP} = \begin{pmatrix} 1 + 2\left(\frac{26}{17}\right) \\ -6 + 3\left(\frac{26}{17}\right) \\ 5 - 2\left(\frac{26}{17}\right) \end{pmatrix} = \begin{pmatrix} 1 + \frac{52}{17} \\ -6 + \frac{78}{17} \\ 5 - \frac{52}{17} \end{pmatrix} = \begin{pmatrix} \frac{17+52}{17} \\ \frac{-102+78}{17} \\ \frac{85-52}{17} \end{pmatrix} = \begin{pmatrix} \frac{69}{17} \\ \frac{-24}{17} \\ \frac{33}{17} \end{pmatrix} $$
6. Calculate the magnitude of \(\vec{CP}\): This is the shortest distance.
$$ |\vec{CP}| = \sqrt{\left(\frac{69}{17}\right)^2 + \left(\frac{-24}{17}\right)^2 + \left(\frac{33}{17}\right)^2} $$
$$ |\vec{CP}| = \sqrt{\frac{4761 + 576 + 1089}{17^2}} = \sqrt{\frac{6426}{289}} = \frac{\sqrt{6426}}{17} $$
The shortest distance is \(\frac{\sqrt{6426}}{17}\) units.

Practice

1. Given points \(A(3, -1, 2)\) and \(B(1, 4, -3)\), find the position vector of the midpoint of \(AB\).
2. A line \(L_1\) has the equation \(\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix}\). Find the coordinates of the point where \(L_1\) intersects the plane \(x=7\).
3. Two lines are given by \(\mathbf{r}_1 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\) and \(\mathbf{r}_2 = \begin{pmatrix} 0 \\ 1 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\). Determine if these lines intersect, and if so, find the position vector of their intersection point.
4. The points \(P\), \(Q\) and \(R\) have position vectors \(\mathbf{p} = 3\mathbf{i} + \mathbf{j} - \mathbf{k}\), \(\mathbf{q} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\) and \(\mathbf{r} = 5\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) respectively. Find the area of the triangle \(PQR\).

Answers:

1. \(\begin{pmatrix} 2 \\ \frac{3}{2} \\ -\frac{1}{2} \end{pmatrix}\)
2. \((7, -4, 4)\)
3. The lines intersect at \(\begin{pmatrix} 3 \\ 0 \\ 7 \end{pmatrix}\).
4. Working for Q4:
   First, find two vectors forming two sides of the triangle, e.g., \(\vec{PQ}\) and \(\vec{PR}\).
   $$ \vec{PQ} = \mathbf{q} - \mathbf{p} = (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) - (3\mathbf{i} + \mathbf{j} - \mathbf{k}) = -2\mathbf{i} + \mathbf{j} + 3\mathbf{k} $$
   $$ \vec{PR} = \mathbf{r} - \mathbf{p} = (5\mathbf{i} - \mathbf{j} + 3\mathbf{k}) - (3\mathbf{i} + \mathbf{j} - \mathbf{k}) = 2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} $$
   The area of triangle \(PQR\) is given by \(\frac{1}{2} |\vec{PQ} \times \vec{PR}|\).
   Calculate the vector (cross) product \(\vec{PQ} \times \vec{PR}\):
   $$ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 3 \\ 2 & -2 & 4 \end{vmatrix} $$
   $$ = \mathbf{i}((1)(4) - (3)(-2)) - \mathbf{j}((-2)(4) - (3)(2)) + \mathbf{k}((-2)(-2) - (1)(2)) $$
   $$ = \mathbf{i}(4 + 6) - \mathbf{j}(-8 - 6) + \mathbf{k}(4 - 2) $$
   $$ = 10\mathbf{i} + 14\mathbf{j} + 2\mathbf{k} $$
   Now, find the magnitude of this vector:
   $$ |10\mathbf{i} + 14\mathbf{j} + 2\mathbf{k}| = \sqrt{10^2 + 14^2 + 2^2} = \sqrt{100 + 196 + 4} = \sqrt{300} $$
   $$ \sqrt{300} = \sqrt{100 \times 3} = 10\sqrt{3} $$
   Finally, the area of the triangle is:
   $$ \text{Area} = \frac{1}{2} |\vec{PQ} \times \vec{PR}| = \frac{1}{2} (10\sqrt{3}) = 5\sqrt{3} \text{ units}^2 $$

The three mistakes that lose marks

1. Confusing position vectors with displacement vectors: Using a position vector \(\mathbf{a}\) as a direction vector for a line passing through the origin, or using \(\mathbf{a}\) directly in a scalar product when the problem requires a vector between two points.
   • Wrong answer produced: Incorrect line equations, or angles calculated relative to the origin instead of between the intended vectors. For example, finding the angle between \(\mathbf{a}\) and \(\mathbf{b}\) instead of \(\vec{AB}\) and \(\vec{CD}\).
2. Incorrectly applying the scalar product for perpendicularity or angle: Forgetting that \(\mathbf{a} \cdot \mathbf{b} = 0\) implies perpendicularity only if both \(\mathbf{a}\) and \(\mathbf{b}\) are non-zero vectors. Also, using the formula \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\) with position vectors when the angle between lines or planes is required, which might involve direction vectors or normal vectors.
   • Wrong answer produced: Stating two vectors are perpendicular when they are not, or calculating an angle that is not the one requested (e.g., angle between position vectors to points on a line, rather than the angle between the lines themselves).
3. Algebraic errors when solving simultaneous equations for intersection points: Setting up the correct vector equations for intersecting lines or a line and a plane, but then making arithmetic mistakes when solving for \(\lambda\) and \(\mu\) (or other parameters). This often involves sign errors or miscalculation of coefficients.
   • Wrong answer produced: An incorrect intersection point, or incorrectly concluding that lines do not intersect when they do (or vice versa) due to inconsistent parameter values.

30-second recap

Vectors in three dimensions extend 2D concepts by adding a \(z\)-component. Key skills are forming line equations \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\), calculating magnitudes \(|\mathbf{v}| = \sqrt{x^2+y^2+z^2}\), and using the scalar product \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta\) for angles and perpendicularity. Shortest distances often involve setting a scalar product to zero.