Home › Subjects › Mathematics 9709 › Vectors Vectors (syllabus ref 3.7) extend coordinate geometry into two and three dimensions, written as columns, as x i + y j + z k x\mathbf{i}+y\mathbf{j}+z\mathbf{k} x i + y j + z k , or as a directed segment A B → \overrightarrow{AB} A B . Adding, subtracting and scaling vectors has a direct geometric meaning: O B → = O A → + O C → \overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{OC} O B = O A + O C describes a parallelogram, and the midpoint of A B AB A B has position vector 1 2 ( a + b ) \tfrac12(\mathbf{a}+\mathbf{b}) 2 1 ( a + b ) . The magnitude of a vector v = ( x y z ) \mathbf{v}=\begin{pmatrix}x\\y\\z\end{pmatrix} v = x y z is ∣ v ∣ = x 2 + y 2 + z 2 |\mathbf{v}|=\sqrt{x^2+y^2+z^2} ∣ v ∣ = x 2 + y 2 + z 2 , and dividing a vector by its own magnitude produces a unit vector in the same direction.
A line’s vector equation, r = a + t b \mathbf{r}=\mathbf{a}+t\mathbf{b} r = a + t b , describes every point on the line as a fixed position vector a \mathbf{a} a plus a variable multiple of a fixed direction vector b \mathbf{b} b . Given two such lines, solving their equations simultaneously reveals whether they are parallel (proportional direction vectors), intersecting (a consistent solution for both parameters exists), or skew (neither, which is only possible in three dimensions). The scalar product a ⋅ b = ∣ a ∣ ∣ b ∣ cos θ \mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta a ⋅ b = ∣ a ∣∣ b ∣ cos θ measures the angle θ \theta θ between two vectors directly, and is the standard tool for finding angles between lines and the foot of a perpendicular from a point to a line, including in solid shapes such as cuboids and tetrahedra.
Original worked problems below apply these vector techniques to two- and three-dimensional geometry.
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01 Question Question 1 Multiple choice A2 1 mark The vector p = 4 i − j + 8 k \mathbf{p} = 4\mathbf{i} - \mathbf{j} + 8\mathbf{k} p = 4 i − j + 8 k .
Which of the following is the unit vector in the direction of p \mathbf{p} p ?
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02 Question Question 2 Structured A2 7 marks Relative to an origin O O O ,
O A → = a = ( 3 − 1 2 ) , O B → = b = ( 1 4 − 1 ) . \overrightarrow{OA} = \mathbf{a} = \begin{pmatrix}3\\-1\\2\end{pmatrix}, \qquad \overrightarrow{OB} = \mathbf{b} = \begin{pmatrix}1\\4\\-1\end{pmatrix}. O A = a = 3 − 1 2 , O B = b = 1 4 − 1 .
(a) Find A B → \overrightarrow{AB} A B . [2]
(b) Find ∣ A B → ∣ |\overrightarrow{AB}| ∣ A B ∣ , giving your answer in the form n \sqrt{n} n for an integer n n n . [2]
(c) Find angle A O B AOB A O B , the angle between O A → \overrightarrow{OA} O A and O B → \overrightarrow{OB} O B , correct to 1 1 1 decimal place. [3]
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03 Question Question 3 Multiple choice A2 2 marks Lines l 1 l_1 l 1 and l 2 l_2 l 2 have vector equations
l 1 : r = ( 2 1 − 1 ) + s ( 4 − 2 6 ) , l 2 : r = ( 1 − 3 2 ) + t ( − 6 3 − 9 ) . l_1: \mathbf{r} = \begin{pmatrix}2\\1\\-1\end{pmatrix} + s\begin{pmatrix}4\\-2\\6\end{pmatrix}, \qquad l_2: \mathbf{r} = \begin{pmatrix}1\\-3\\2\end{pmatrix} + t\begin{pmatrix}-6\\3\\-9\end{pmatrix}. l 1 : r = 2 1 − 1 + s 4 − 2 6 , l 2 : r = 1 − 3 2 + t − 6 3 − 9 .
What is the relationship between l 1 l_1 l 1 and l 2 l_2 l 2 ?
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04 Question Question 4 Structured A2 9 marks Lines l 1 l_1 l 1 and l 2 l_2 l 2 have vector equations
l 1 : r = ( 1 2 − 3 ) + s ( 2 − 1 1 ) , l 2 : r = ( 1 − 1 2 ) + t ( 1 1 − 2 ) . l_1: \mathbf{r} = \begin{pmatrix}1\\2\\-3\end{pmatrix} + s\begin{pmatrix}2\\-1\\1\end{pmatrix}, \qquad l_2: \mathbf{r} = \begin{pmatrix}1\\-1\\2\end{pmatrix} + t\begin{pmatrix}1\\1\\-2\end{pmatrix}. l 1 : r = 1 2 − 3 + s 2 − 1 1 , l 2 : r = 1 − 1 2 + t 1 1 − 2 .
(a) Show that l 1 l_1 l 1 and l 2 l_2 l 2 are not parallel. [2]
(b) Show that l 1 l_1 l 1 and l 2 l_2 l 2 intersect, and find the position vector of their point of intersection. [4]
(c) Find the acute angle between l 1 l_1 l 1 and l 2 l_2 l 2 , correct to 1 1 1 decimal place. [3]
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05 Question Question 5 Structured A2 8 marks Points P P P and Q Q Q have position vectors O P → = ( 2 − 1 5 ) \overrightarrow{OP} = \begin{pmatrix}2\\-1\\5\end{pmatrix} O P = 2 − 1 5 and O Q → = ( 6 3 − 3 ) \overrightarrow{OQ} = \begin{pmatrix}6\\3\\-3\end{pmatrix} O Q = 6 3 − 3 .
(a) Find a vector equation for the line P Q PQ P Q , giving the direction vector in its simplest integer form. [3]
(b) The point R ( − 2 , − 5 , 13 ) R(-2,-5,13) R ( − 2 , − 5 , 13 ) is claimed to lie on line P Q PQ P Q . Determine, showing full working, whether this is true. [3]
(c) Find the unit vector in the direction of P Q → \overrightarrow{PQ} P Q . [2]
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06 Question Question 6 Multiple choice A2 1 mark The vector v = 2 i − 3 j + 6 k \mathbf{v} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k} v = 2 i − 3 j + 6 k .
What is ∣ v ∣ |\mathbf{v}| ∣ v ∣ ?
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07 Question Question 7 Structured A2 7 marks Points M M M and N N N have position vectors
O M → = ( 2 5 − 3 ) , O N → = ( 5 9 9 ) \overrightarrow{OM} = \begin{pmatrix}2\\5\\-3\end{pmatrix}, \qquad \overrightarrow{ON} = \begin{pmatrix}5\\9\\9\end{pmatrix} O M = 2 5 − 3 , O N = 5 9 9
relative to an origin O O O .
(a) Find M N → \overrightarrow{MN} M N . [2]
(b) Find ∣ M N → ∣ |\overrightarrow{MN}| ∣ M N ∣ . [2]
(c) Find the unit vector in the direction of M N → \overrightarrow{MN} M N , and hence write down the unit vector in the direction of N M → \overrightarrow{NM} N M . [3]
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08 Question Question 8 Multiple choice A2 1 mark Vectors a = ( 1 2 2 ) \mathbf{a} = \begin{pmatrix}1\\2\\2\end{pmatrix} a = 1 2 2 and b = ( 2 − 2 1 ) \mathbf{b} = \begin{pmatrix}2\\-2\\1\end{pmatrix} b = 2 − 2 1 .
What is the angle between a \mathbf{a} a and b \mathbf{b} b ?
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09 Question Question 9 Structured A2 7 marks Points A A A , B B B and C C C have position vectors
O A → = ( 1 − 2 3 ) , O B → = ( 3 2 1 ) , O C → = ( 4 − 2 7 ) \overrightarrow{OA} = \begin{pmatrix}1\\-2\\3\end{pmatrix}, \qquad \overrightarrow{OB} = \begin{pmatrix}3\\2\\1\end{pmatrix}, \qquad \overrightarrow{OC} = \begin{pmatrix}4\\-2\\7\end{pmatrix} O A = 1 − 2 3 , O B = 3 2 1 , O C = 4 − 2 7
relative to an origin O O O .
(a) Find A B → \overrightarrow{AB} A B and A C → \overrightarrow{AC} A C . [2]
(b) Use the scalar product to find angle B A C BAC B A C , correct to 1 1 1 decimal place. [3]
(c) State, with a reason, whether angle B A C BAC B A C is exactly 90 ∘ 90^\circ 9 0 ∘ . [2]
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10 Question Question 10 Structured A2 7 marks Vectors a = ( 3 k − 2 ) \mathbf{a} = \begin{pmatrix}3\\k\\-2\end{pmatrix} a = 3 k − 2 and b = ( 2 − 1 4 ) \mathbf{b} = \begin{pmatrix}2\\-1\\4\end{pmatrix} b = 2 − 1 4 , where k k k is a constant.
(a) Given that a \mathbf{a} a and b \mathbf{b} b are perpendicular, find the value of k k k . [3]
(b) Using this value of k k k , find ∣ a ∣ |\mathbf{a}| ∣ a ∣ , giving your answer in the form n \sqrt{n} n for an integer n n n . [2]
(c) Find the unit vector in the direction of a \mathbf{a} a . [2]
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