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⬅ Intersection of two lines
Displacement or translation vectors Position vectors The scalar product The vector equation of a line Intersection of two lines Review

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Vectors — Review

This review brings together the main Year 13 Cambridge 9709 vector ideas in three dimensions: displacement vectors, position vectors, scalar product, equations of lines, and intersection of two lines.

In exam questions, success usually depends on choosing the correct formula, writing vectors clearly, and checking coordinates carefully.

Key definitions and formulae

1. Displacement and translation vectors

If a point \(P(x,y,z)\) is translated by

\[ \begin{pmatrix} a\\ b\\ c \end{pmatrix}, \]

then its image is

\[ P'(x+a,\;y+b,\;z+c). \]

If \(A(x_1,y_1,z_1)\) and \(B(x_2,y_2,z_2)\), then

\[ \overrightarrow{AB}= \begin{pmatrix} x_2-x_1\\ y_2-y_1\\ z_2-z_1 \end{pmatrix}. \]

2. Position vectors

If \(P(x,y,z)\), then the position vector of \(P\) is

\[ \overrightarrow{OP}= \begin{pmatrix} x\\ y\\ z \end{pmatrix}. \]

If points \(A\) and \(B\) have position vectors \(\mathbf{a}\) and \(\mathbf{b}\), then

\[ \overrightarrow{AB}=\mathbf{b}-\mathbf{a}. \]

The midpoint \(M\) of \(AB\) has position vector

\[ \overrightarrow{OM}=\frac{\mathbf{a}+\mathbf{b}}{2}. \]

3. Scalar product

If

\[ \mathbf{a}= \begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix}, \qquad \mathbf{b}= \begin{pmatrix} b_1\\ b_2\\ b_3 \end{pmatrix}, \]

then

\[ \mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3. \]

In \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form:

\[ \mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}, \qquad \mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}. \]

Also,

\[ \mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta. \]

Two non-zero vectors are perpendicular if

\[ \mathbf{a}\cdot\mathbf{b}=0. \]

4. Magnitude of a vector

\[ \left| \begin{pmatrix} a\\ b\\ c \end{pmatrix} \right| =\sqrt{a^2+b^2+c^2}. \]

5. Vector equation of a line

A line through the point \(A(x_1,y_1,z_1)\) with direction vector

\[ \begin{pmatrix} a\\ b\\ c \end{pmatrix} \]

can be written in column vector form as

\[ \mathbf{r}= \begin{pmatrix} x_1\\ y_1\\ z_1 \end{pmatrix} + \lambda \begin{pmatrix} a\\ b\\ c \end{pmatrix}. \]

Parametric form:

\[ x=x_1+a\lambda,\qquad y=y_1+b\lambda,\qquad z=z_1+c\lambda. \]

\( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form:

\[ \mathbf{r}=(x_1\mathbf{i}+y_1\mathbf{j}+z_1\mathbf{k}) + \lambda(a\mathbf{i}+b\mathbf{j}+c\mathbf{k}). \]

6. Intersection of two lines

If

\[ \mathbf{r}=\mathbf{a}+\lambda\mathbf{b} \qquad \text{and} \qquad \mathbf{r}=\mathbf{c}+\mu\mathbf{d}, \]

then to test for intersection, set them equal:

\[ \mathbf{a}+\lambda\mathbf{b}=\mathbf{c}+\mu\mathbf{d}. \]

Compare the three coordinates and solve for \( \lambda \) and \( \mu \). The lines intersect only if all three coordinates are consistent.

Worked examples

Example 1: Displacement vector

Find the displacement vector from \(A(1,-2,4)\) to \(B(5,3,-1)\).

Solution

\[ \overrightarrow{AB}= \begin{pmatrix} 5-1\\ 3-(-2)\\ -1-4 \end{pmatrix} = \begin{pmatrix} 4\\ 5\\ -5 \end{pmatrix}. \]
So the displacement vector is \[ \begin{pmatrix} 4\\ 5\\ -5 \end{pmatrix}. \]

Example 2: Position vector and midpoint

The position vectors of \(A\) and \(B\) are

\[ \mathbf{a}= \begin{pmatrix} 2\\ 1\\ -3 \end{pmatrix}, \qquad \mathbf{b}= \begin{pmatrix} 6\\ -5\\ 7 \end{pmatrix}. \]

Find the position vector of the midpoint \(M\).

Solution

\[ \overrightarrow{OM}=\frac{\mathbf{a}+\mathbf{b}}{2} = \frac{1}{2} \begin{pmatrix} 8\\ -4\\ 4 \end{pmatrix} = \begin{pmatrix} 4\\ -2\\ 2 \end{pmatrix}. \]
The midpoint has position vector \[ \begin{pmatrix} 4\\ -2\\ 2 \end{pmatrix}. \]

Example 3: Scalar product in \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form

Find \( \mathbf{p}\cdot\mathbf{q} \) if

\[ \mathbf{p}=2\mathbf{i}-\mathbf{j}+3\mathbf{k}, \qquad \mathbf{q}=4\mathbf{i}+5\mathbf{j}-2\mathbf{k}. \]

Solution

\[ \mathbf{p}\cdot\mathbf{q}=(2)(4)+(-1)(5)+(3)(-2)=8-5-6=-3. \]
Therefore, \[ \mathbf{p}\cdot\mathbf{q}=-3. \]

Example 4: Equation of a line

Write the equation of the line through \(A(1,2,-3)\) with direction vector

\[ \begin{pmatrix} 4\\ -3\\ 2 \end{pmatrix}. \]

Solution

Column vector form:

\[ \mathbf{r}= \begin{pmatrix} 1\\ 2\\ -3 \end{pmatrix} + \lambda \begin{pmatrix} 4\\ -3\\ 2 \end{pmatrix}. \]

Parametric form:

\[ x=1+4\lambda,\qquad y=2-3\lambda,\qquad z=-3+2\lambda. \]

\( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form:

\[ \mathbf{r}=(\mathbf{i}+2\mathbf{j}-3\mathbf{k}) + \lambda(4\mathbf{i}-3\mathbf{j}+2\mathbf{k}). \]
All three forms represent the same line.

Example 5: Intersection of two lines

Find whether the lines

\[ \mathbf{r}= \begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix} \] \[ \mathbf{r}= \begin{pmatrix} 5\\ 0\\ 11 \end{pmatrix} + \mu \begin{pmatrix} -1\\ 1\\ -2 \end{pmatrix} \]

intersect.

Solution

\[ \begin{pmatrix} 1+2\lambda\\ 2-\lambda\\ 3+4\lambda \end{pmatrix} = \begin{pmatrix} 5-\mu\\ \mu\\ 11-2\mu \end{pmatrix}. \]

Comparing coordinates gives

\[ 2-\lambda=\mu. \]

Substitute into \(1+2\lambda=5-\mu\):

\[ 1+2\lambda=5-(2-\lambda)=3+\lambda \qquad \Rightarrow \qquad \lambda=2. \]

Then

\[ \mu=2-\lambda=0. \]

Check the third coordinate:

\[ 3+4(2)=11-2(0) \qquad \Rightarrow \qquad 11=11. \]
The lines intersect at \[ \begin{pmatrix} 5\\ 0\\ 11 \end{pmatrix}. \]

Standard methods

Method 1: Vector from one point to another

\[ \text{finish} - \text{start}. \]

Method 2: Scalar product

Multiply matching components and add.

\[ \mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3. \]

Method 3: Angle between two vectors

\[ \cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}. \]

Method 4: Equation of a line

Use one point on the line and one direction vector.

\[ \mathbf{r}=\mathbf{a}+\lambda\mathbf{b}. \]

Method 5: Intersection of two lines

Set the lines equal, solve for the parameters, and check all three coordinates.

\[ \mathbf{a}+\lambda\mathbf{b}=\mathbf{c}+\mu\mathbf{d}. \]

Common mistakes and exam tips

Common mistakes

  • Reversing subtraction when finding a displacement vector.
  • Confusing a point such as \((x,y,z)\) with a vector such as \(\begin{pmatrix}x\\y\\z\end{pmatrix}\).
  • Thinking the scalar product gives a vector. It gives a scalar.
  • Forgetting to find magnitudes when using the angle formula.
  • Using only two coordinates when checking whether two lines intersect.
  • Mixing up column vector form, parametric form, and \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form.

Exam tips

  • Write vectors clearly as column vectors or in \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form.
  • Use finish minus start for displacement vectors.
  • For perpendicular vectors, check whether \( \mathbf{a}\cdot\mathbf{b}=0 \).
  • For a line, always identify the point and the direction vector.
  • When testing intersection, always check all three coordinates before giving your conclusion.

Summary

\[ \overrightarrow{AB}=\mathbf{b}-\mathbf{a} \] \[ \mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3 \] \[ \mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta \] \[ \mathbf{r}=\mathbf{a}+\lambda\mathbf{b} \] \[ \mathbf{a}+\lambda\mathbf{b}=\mathbf{c}+\mu\mathbf{d} \]

Year 13 vectors combine algebra and geometry. The key ideas are describing positions, finding vectors between points, using the scalar product, writing equations of lines, and checking whether lines meet.

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