Further Maths Vector and triple products Practice Questions

Free Further Maths Vector and triple products practice questions with full step-by-step worked solutions. Covers vector-product, cross-product, anticommutativity, magnitude. Practise exam-style problems and check your method.

vector-productcross-productanticommutativitymagnitudeareaparallelogram
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Given a=(123)\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} and b=(456)\mathbf{b} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}, find a×b\mathbf{a} \times \mathbf{b}.
Show worked solution

Worked solution

  1. Set up the determinant for the vector product

    a×b=∣ijk123456∣\mathbf{a}\times \mathbf{b}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix}

    Place i,j,k\mathbf{i},\mathbf{j},\mathbf{k} on the top row and the components beneath them.

  2. Expand the i\mathbf{i} component

    (2)(6)−(3)(5)=−3(2)(6)-(3)(5)=-3

    The i\mathbf{i} component is a2b3−a3b2a_2b_3-a_3b_2.

  3. Expand the j\mathbf{j} component, remembering the minus sign

    −[(1)(6)−(3)(4)]=6-\big[(1)(6)-(3)(4)\big]=6

    The cofactor expansion attaches a minus sign to the j\mathbf{j} term, giving a3b1−a1b3a_3b_1-a_1b_3.

  4. State the vector product

    a×b=(−36−3)\mathbf{a}\times\mathbf{b}=\begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix}

    This vector is perpendicular to both of the given vectors.

Answer
(−36−3)\begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix}
Question 2
2 markseasy
The vectors a=(300)\mathbf{a} = \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix}, b=(020)\mathbf{b} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} and c=(115)\mathbf{c} = \begin{pmatrix} 1 \\ 1 \\ 5 \end{pmatrix} form three edges of a parallelepiped. Which of the following is the volume of the parallelepiped?
Show worked solution

Worked solution

  1. Form the vector product b×c\mathbf{b}\times \mathbf{c}

    b×c=∣ijk020115∣=(100−2)\mathbf{b}\times \mathbf{c}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 0 \\ 1 & 1 & 5 \end{vmatrix}=\begin{pmatrix} 10 \\ 0 \\ -2 \end{pmatrix}

    Expand the determinant to obtain a vector normal to b\mathbf{b} and c\mathbf{c}.

  2. Evaluate the scalar triple product

    a⋅(b×c)=30\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})=30

    This single number is the signed volume of the parallelepiped.

  3. Take the modulus

    V=∣30∣=30V=\left|30\right|=30

    Volume cannot be negative.

  4. Select the correct volume

    3030

    The volume is the modulus of the scalar triple product.

Answer
3030
Question 3
4 marksintermediate
Given a=(1−24)\mathbf{a} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} and b=(35−1)\mathbf{b} = \begin{pmatrix} 3 \\ 5 \\ -1 \end{pmatrix}, which of the following is equal to b×a\mathbf{b} \times \mathbf{a}?
Show worked solution

Worked solution

  1. Compute a×b\mathbf{a}\times\mathbf{b} first

    a×b=∣ijk1−2435−1∣=(−181311)\mathbf{a}\times\mathbf{b}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & 4 \\ 3 & 5 & -1 \end{vmatrix}=\begin{pmatrix} -18 \\ 13 \\ 11 \end{pmatrix}

    Expand the determinant in the given order.

  2. Apply the anticommutative property

    b×a=−(a×b)=(18−13−11)\mathbf{b}\times\mathbf{a}=-\left(\mathbf{a}\times\mathbf{b}\right)=\begin{pmatrix} 18 \\ -13 \\ -11 \end{pmatrix}

    Swapping the two vectors reverses the sign of the result.

  3. Set up the determinant for the vector product

    b×a=∣ijk35−11−24∣\mathbf{b}\times \mathbf{a}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 5 & -1 \\ 1 & -2 & 4 \end{vmatrix}

    Place i,j,k\mathbf{i},\mathbf{j},\mathbf{k} on the top row and the components beneath them.

  4. Expand the i\mathbf{i} component

    (5)(4)−(−1)(−2)=18(5)(4)-(-1)(-2)=18

    The i\mathbf{i} component is a2b3−a3b2a_2b_3-a_3b_2.

  5. Expand the j\mathbf{j} component, remembering the minus sign

    −[(3)(4)−(−1)(1)]=−13-\big[(3)(4)-(-1)(1)\big]=-13

    The cofactor expansion attaches a minus sign to the j\mathbf{j} term, giving a3b1−a1b3a_3b_1-a_1b_3.

  6. Expand the k\mathbf{k} component

    (3)(−2)−(5)(1)=−11(3)(-2)-(5)(1)=-11

    The k\mathbf{k} component is a1b2−a2b1a_1b_2-a_2b_1.

  7. Select the correct vector

    (18−13−11)\begin{pmatrix} 18 \\ -13 \\ -11 \end{pmatrix}

    Reversing the order negates every component of the vector product.

Answer
(18−13−11)\begin{pmatrix} 18 \\ -13 \\ -11 \end{pmatrix}
Question 4
6 markshard
Given a=(4−17)\mathbf{a} = \begin{pmatrix} 4 \\ -1 \\ 7 \end{pmatrix} and b=(23−5)\mathbf{b} = \begin{pmatrix} 2 \\ 3 \\ -5 \end{pmatrix}, which of the following is equal to b×a\mathbf{b} \times \mathbf{a}?
Show worked solution

Worked solution

  1. Compute a×b\mathbf{a}\times\mathbf{b} first

    a×b=∣ijk4−1723−5∣=(−163414)\mathbf{a}\times\mathbf{b}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 7 \\ 2 & 3 & -5 \end{vmatrix}=\begin{pmatrix} -16 \\ 34 \\ 14 \end{pmatrix}

    Expand the determinant in the given order.

  2. Apply the anticommutative property

    b×a=−(a×b)=(16−34−14)\mathbf{b}\times\mathbf{a}=-\left(\mathbf{a}\times\mathbf{b}\right)=\begin{pmatrix} 16 \\ -34 \\ -14 \end{pmatrix}

    Swapping the two vectors reverses the sign of the result.

  3. Set up the determinant for the vector product

    b×a=∣ijk23−54−17∣\mathbf{b}\times \mathbf{a}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & -5 \\ 4 & -1 & 7 \end{vmatrix}

    Place i,j,k\mathbf{i},\mathbf{j},\mathbf{k} on the top row and the components beneath them.

  4. Expand the i\mathbf{i} component

    (3)(7)−(−5)(−1)=16(3)(7)-(-5)(-1)=16

    The i\mathbf{i} component is a2b3−a3b2a_2b_3-a_3b_2.

  5. Expand the j\mathbf{j} component, remembering the minus sign

    −[(2)(7)−(−5)(4)]=−34-\big[(2)(7)-(-5)(4)\big]=-34

    The cofactor expansion attaches a minus sign to the j\mathbf{j} term, giving a3b1−a1b3a_3b_1-a_1b_3.

  6. Expand the k\mathbf{k} component

    (2)(−1)−(3)(4)=−14(2)(-1)-(3)(4)=-14

    The k\mathbf{k} component is a1b2−a2b1a_1b_2-a_2b_1.

  7. Assemble the vector product

    b×a=(16−34−14)\mathbf{b}\times \mathbf{a}=\begin{pmatrix} 16 \\ -34 \\ -14 \end{pmatrix}

    Collect the three components into a single column vector.

  8. Check the result is perpendicular to b\mathbf{b}

    b⋅(b×a)=(2)(16)+(3)(−34)+(−5)(−14)=0\mathbf{b}\cdot(\mathbf{b}\times \mathbf{a})=(2)(16)+(3)(-34)+(-5)(-14)=0

    A zero scalar product confirms the vector product is normal to b\mathbf{b}.

  9. Check the result is perpendicular to a\mathbf{a}

    a⋅(b×a)=(4)(16)+(−1)(−34)+(7)(−14)=0\mathbf{a}\cdot(\mathbf{b}\times \mathbf{a})=(4)(16)+(-1)(-34)+(7)(-14)=0

    A second zero scalar product confirms the direction is normal to the whole plane.

  10. Confirm the anticommutative property

    a×b=(−163414)\mathbf{a}\times \mathbf{b}=\begin{pmatrix} -16 \\ 34 \\ 14 \end{pmatrix}

    Reversing the order of the vectors reverses every component.

  11. Select the correct vector

    (16−34−14)\begin{pmatrix} 16 \\ -34 \\ -14 \end{pmatrix}

    Reversing the order negates every component of the vector product.

Answer
(16−34−14)\begin{pmatrix} 16 \\ -34 \\ -14 \end{pmatrix}
Question 5
9 markschallenging
Given a=(73−5)\mathbf{a} = \begin{pmatrix} 7 \\ 3 \\ -5 \end{pmatrix} and b=(−264)\mathbf{b} = \begin{pmatrix} -2 \\ 6 \\ 4 \end{pmatrix}, which of the following is the exact value of ∣a×b∣\left|\mathbf{a} \times \mathbf{b}\right|?
Show worked solution

Worked solution

  1. Form the vector product

    a×b=∣ijk73−5−264∣=(42−1848)\mathbf{a}\times \mathbf{b}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 7 & 3 & -5 \\ -2 & 6 & 4 \end{vmatrix}=\begin{pmatrix} 42 \\ -18 \\ 48 \end{pmatrix}

    Expand the determinant to find the vector normal to both.

  2. Square each component

    422+−182+482=439242^2+-18^2+48^2=4392

    The magnitude formula needs the sum of the squares of the components.

  3. Take the square root

    ∣a×b∣=4392=6122\left|\mathbf{a}\times \mathbf{b}\right|=\sqrt{4392}=6\sqrt{122}

    Simplify the surd by extracting any square factors.

  4. Expand the i\mathbf{i} component

    (3)(4)−(−5)(6)=42(3)(4)-(-5)(6)=42

    The i\mathbf{i} component is a2b3−a3b2a_2b_3-a_3b_2.

  5. Expand the j\mathbf{j} component, remembering the minus sign

    −[(7)(4)−(−5)(−2)]=−18-\big[(7)(4)-(-5)(-2)\big]=-18

    The cofactor expansion attaches a minus sign to the j\mathbf{j} term, giving a3b1−a1b3a_3b_1-a_1b_3.

  6. Expand the k\mathbf{k} component

    (7)(6)−(3)(−2)=48(7)(6)-(3)(-2)=48

    The k\mathbf{k} component is a1b2−a2b1a_1b_2-a_2b_1.

  7. Check the result is perpendicular to a\mathbf{a}

    a⋅(a×b)=(7)(42)+(3)(−18)+(−5)(48)=0\mathbf{a}\cdot(\mathbf{a}\times \mathbf{b})=(7)(42)+(3)(-18)+(-5)(48)=0

    A zero scalar product confirms the vector product is normal to a\mathbf{a}.

  8. Check the result is perpendicular to b\mathbf{b}

    b⋅(a×b)=(−2)(42)+(6)(−18)+(4)(48)=0\mathbf{b}\cdot(\mathbf{a}\times \mathbf{b})=(-2)(42)+(6)(-18)+(4)(48)=0

    A second zero scalar product confirms the direction is normal to the whole plane.

  9. Confirm the anticommutative property

    b×a=(−4218−48)\mathbf{b}\times \mathbf{a}=\begin{pmatrix} -42 \\ 18 \\ -48 \end{pmatrix}

    Reversing the order of the vectors reverses every component.

  10. Find the magnitude of the vector product

    ∣a×b∣=422+−182+482=4392=6122\left|\mathbf{a}\times \mathbf{b}\right|=\sqrt{42^2+-18^2+48^2}=\sqrt{4392}=6\sqrt{122}

    The magnitude is the area of the parallelogram spanned by the two vectors.

  11. Interpret the magnitude geometrically

    Area of parallelogram=6122\text{Area of parallelogram}=6\sqrt{122}

    The length of a×b\mathbf{a}\times \mathbf{b} measures the area swept out by the two vectors.

  12. Note the scalar product of the two vectors

    a⋅b=−16\mathbf{a}\cdot \mathbf{b}=-16

    A non-zero scalar product shows the vectors are not perpendicular to each other.

  13. Recall the definition of the vector product

    a×b=∣a∣∣b∣sin⁡θ n^\mathbf{a}\times\mathbf{b}=\left|\mathbf{a}\right|\left|\mathbf{b}\right|\sin\theta\,\hat{\mathbf{n}}

    The vector product has magnitude ∣a∣∣b∣sin⁡θ|\mathbf{a}||\mathbf{b}|\sin\theta and direction given by the right-hand rule.

  14. Recall that the vector product is anticommutative

    a×b=−(b×a)\mathbf{a}\times\mathbf{b}=-\left(\mathbf{b}\times\mathbf{a}\right)

    Swapping the order of the two vectors reverses the sign of every component.

  15. Select the correct magnitude

    61226\sqrt{122}

    This is the exact magnitude of the vector product.

Answer
61226\sqrt{122}

Unlock 65 more Vector and triple products questions

Create a free account to work through every Further Maths Vector and triple products question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Vector and triple products practice

Hard QuestionsWorked Solutions

Related Further Pure topics

Conic sections 1Conic sections 2Further complex numbersGroupsInequalitiesFurther integration techniques