Hard Further Maths Vector and triple products Questions

Challenging, exam-style Further Maths Vector and triple products questions with worked solutions. Stretch yourself on the hardest vector-product, plane, cartesian-equation, normal problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
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}
Question 2
9 markschallenging
The vectors a=(5−62)\mathbf{a} = \begin{pmatrix} 5 \\ -6 \\ 2 \end{pmatrix} and b=(34−7)\mathbf{b} = \begin{pmatrix} 3 \\ 4 \\ -7 \end{pmatrix} are adjacent sides of a parallelogram. Which of the following is the exact area of the parallelogram?
Show worked solution

Worked solution

  1. Recall the area formula

    Area=∣a×b∣\text{Area}=\left|\mathbf{a}\times\mathbf{b}\right|

    The two vectors are adjacent sides of the parallelogram.

  2. Form the vector product

    a×b=∣ijk5−6234−7∣=(344138)\mathbf{a}\times \mathbf{b}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & -6 & 2 \\ 3 & 4 & -7 \end{vmatrix}=\begin{pmatrix} 34 \\ 41 \\ 38 \end{pmatrix}

    Expand the determinant to find the vector normal to both.

  3. Square each component

    342+412+382=428134^2+41^2+38^2=4281

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

  4. Take the square root

    ∣a×b∣=4281=4281\left|\mathbf{a}\times \mathbf{b}\right|=\sqrt{4281}=\sqrt{4281}

    Simplify the surd by extracting any square factors.

  5. Expand the i\mathbf{i} component

    (−6)(−7)−(2)(4)=34(-6)(-7)-(2)(4)=34

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

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

    −[(5)(−7)−(2)(3)]=41-\big[(5)(-7)-(2)(3)\big]=41

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

  7. Expand the k\mathbf{k} component

    (5)(4)−(−6)(3)=38(5)(4)-(-6)(3)=38

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

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

    a⋅(a×b)=(5)(34)+(−6)(41)+(2)(38)=0\mathbf{a}\cdot(\mathbf{a}\times \mathbf{b})=(5)(34)+(-6)(41)+(2)(38)=0

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

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

    b⋅(a×b)=(3)(34)+(4)(41)+(−7)(38)=0\mathbf{b}\cdot(\mathbf{a}\times \mathbf{b})=(3)(34)+(4)(41)+(-7)(38)=0

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

  10. Confirm the anticommutative property

    b×a=(−34−41−38)\mathbf{b}\times \mathbf{a}=\begin{pmatrix} -34 \\ -41 \\ -38 \end{pmatrix}

    Reversing the order of the vectors reverses every component.

  11. Find the magnitude of the vector product

    ∣a×b∣=342+412+382=4281=4281\left|\mathbf{a}\times \mathbf{b}\right|=\sqrt{34^2+41^2+38^2}=\sqrt{4281}=\sqrt{4281}

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

  12. Interpret the magnitude geometrically

    Area of parallelogram=4281\text{Area of parallelogram}=\sqrt{4281}

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

  13. Note the scalar product of the two vectors

    a⋅b=−23\mathbf{a}\cdot \mathbf{b}=-23

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

  14. 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.

  15. Select the correct area

    4281\sqrt{4281}

    The area of the parallelogram is the magnitude of the vector product.

Answer
4281\sqrt{4281}
Question 3
9 markschallenging
The vectors a=(47−2)\mathbf{a} = \begin{pmatrix} 4 \\ 7 \\ -2 \end{pmatrix}, b=(−315)\mathbf{b} = \begin{pmatrix} -3 \\ 1 \\ 5 \end{pmatrix} and c=(6−43)\mathbf{c} = \begin{pmatrix} 6 \\ -4 \\ 3 \end{pmatrix} form three edges of a parallelepiped. Which of the following is the volume of the parallelepiped?
Show worked solution

Worked solution

  1. Write down the three vectors

    a=(47−2),b=(−315),c=(6−43)\mathbf{a}=\begin{pmatrix} 4 \\ 7 \\ -2 \end{pmatrix},\quad \mathbf{b}=\begin{pmatrix} -3 \\ 1 \\ 5 \end{pmatrix},\quad \mathbf{c}=\begin{pmatrix} 6 \\ -4 \\ 3 \end{pmatrix}

    The scalar triple product needs all three sets of components.

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

    b×c=∣ijk−3156−43∣=(23396)\mathbf{b}\times \mathbf{c}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 1 & 5 \\ 6 & -4 & 3 \end{vmatrix}=\begin{pmatrix} 23 \\ 39 \\ 6 \end{pmatrix}

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

  3. Take the scalar product with a\mathbf{a}

    a⋅(b×c)=(4)(23)+(7)(39)+(−2)(6)\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})=(4)(23)+(7)(39)+(-2)(6)

    Multiply matching components and add.

  4. Evaluate the scalar triple product

    a⋅(b×c)=353\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})=353

    This single number is the signed volume of the parallelepiped.

  5. Take the modulus

    V=∣353∣=353V=\left|353\right|=353

    Volume cannot be negative.

  6. Write the triple product as a determinant

    a⋅(b×c)=∣47−2−3156−43∣\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})=\begin{vmatrix} 4 & 7 & -2 \\ -3 & 1 & 5 \\ 6 & -4 & 3 \end{vmatrix}

    The rows of the determinant are the three vectors in order.

  7. Expand the determinant along the first row

    4∣15−43∣−7∣−3563∣+−2∣−316−4∣4\begin{vmatrix} 1 & 5 \\ -4 & 3 \end{vmatrix}-7\begin{vmatrix} -3 & 5 \\ 6 & 3 \end{vmatrix}+-2\begin{vmatrix} -3 & 1 \\ 6 & -4 \end{vmatrix}

    The three 2×22\times 2 minors give the same value as the two-product method.

  8. Evaluate the three minors

    4(23)−7(−39)+−2(6)=3534(23)-7(-39)+-2(6)=353

    Each minor is a 2×22\times 2 determinant of the remaining components.

  9. Check the cyclic property

    b⋅(c×a)=353\mathbf{b}\cdot(\mathbf{c}\times \mathbf{a})=353

    Cycling the vectors leaves the scalar triple product unchanged.

  10. Check the effect of swapping two vectors

    b⋅(a×c)=−353\mathbf{b}\cdot(\mathbf{a}\times \mathbf{c})=-353

    Interchanging two of the vectors reverses the sign.

  11. Note the sign of the result

    sign⁡=+\operatorname{sign}=+

    A positive value means the three vectors form a right-handed set.

  12. Expand the i\mathbf{i} component

    (1)(3)−(5)(−4)=23(1)(3)-(5)(-4)=23

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

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

    −[(−3)(3)−(5)(6)]=39-\big[(-3)(3)-(5)(6)\big]=39

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

  14. Expand the k\mathbf{k} component

    (−3)(−4)−(1)(6)=6(-3)(-4)-(1)(6)=6

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

  15. Select the correct volume

    353353

    The volume is the modulus of the scalar triple product.

Answer
353353
Question 4
9 markschallenging
Given a=(5−32)\mathbf{a} = \begin{pmatrix} 5 \\ -3 \\ 2 \end{pmatrix}, b=(16−4)\mathbf{b} = \begin{pmatrix} 1 \\ 6 \\ -4 \end{pmatrix} and c=(−217)\mathbf{c} = \begin{pmatrix} -2 \\ 1 \\ 7 \end{pmatrix}, which of the following is the value of a⋅(b×c)\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})?
Show worked solution

Worked solution

  1. Write down the three vectors

    a=(5−32),b=(16−4),c=(−217)\mathbf{a}=\begin{pmatrix} 5 \\ -3 \\ 2 \end{pmatrix},\quad \mathbf{b}=\begin{pmatrix} 1 \\ 6 \\ -4 \end{pmatrix},\quad \mathbf{c}=\begin{pmatrix} -2 \\ 1 \\ 7 \end{pmatrix}

    The scalar triple product needs all three sets of components.

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

    b×c=∣ijk16−4−217∣=(46113)\mathbf{b}\times \mathbf{c}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 6 & -4 \\ -2 & 1 & 7 \end{vmatrix}=\begin{pmatrix} 46 \\ 1 \\ 13 \end{pmatrix}

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

  3. Take the scalar product with a\mathbf{a}

    a⋅(b×c)=(5)(46)+(−3)(1)+(2)(13)\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})=(5)(46)+(-3)(1)+(2)(13)

    Multiply matching components and add.

  4. Evaluate the scalar triple product

    a⋅(b×c)=253\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})=253

    This single number is the signed volume of the parallelepiped.

  5. Write the triple product as a determinant

    a⋅(b×c)=∣5−3216−4−217∣\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})=\begin{vmatrix} 5 & -3 & 2 \\ 1 & 6 & -4 \\ -2 & 1 & 7 \end{vmatrix}

    The rows of the determinant are the three vectors in order.

  6. Expand the determinant along the first row

    5∣6−417∣−−3∣1−4−27∣+2∣16−21∣5\begin{vmatrix} 6 & -4 \\ 1 & 7 \end{vmatrix}--3\begin{vmatrix} 1 & -4 \\ -2 & 7 \end{vmatrix}+2\begin{vmatrix} 1 & 6 \\ -2 & 1 \end{vmatrix}

    The three 2×22\times 2 minors give the same value as the two-product method.

  7. Evaluate the three minors

    5(46)−−3(−1)+2(13)=2535(46)--3(-1)+2(13)=253

    Each minor is a 2×22\times 2 determinant of the remaining components.

  8. Check the cyclic property

    b⋅(c×a)=253\mathbf{b}\cdot(\mathbf{c}\times \mathbf{a})=253

    Cycling the vectors leaves the scalar triple product unchanged.

  9. Check the effect of swapping two vectors

    b⋅(a×c)=−253\mathbf{b}\cdot(\mathbf{a}\times \mathbf{c})=-253

    Interchanging two of the vectors reverses the sign.

  10. Note the sign of the result

    sign⁡=+\operatorname{sign}=+

    A positive value means the three vectors form a right-handed set.

  11. Expand the i\mathbf{i} component

    (6)(7)−(−4)(1)=46(6)(7)-(-4)(1)=46

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

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

    −[(1)(7)−(−4)(−2)]=1-\big[(1)(7)-(-4)(-2)\big]=1

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

  13. Expand the k\mathbf{k} component

    (1)(1)−(6)(−2)=13(1)(1)-(6)(-2)=13

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

  14. 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.

  15. Select the correct value

    253253

    This is the value of the scalar triple product.

Answer
253253
Question 5
9 markschallenging
Given a=(6−52)\mathbf{a} = \begin{pmatrix} 6 \\ -5 \\ 2 \end{pmatrix} and b=(−437)\mathbf{b} = \begin{pmatrix} -4 \\ 3 \\ 7 \end{pmatrix}, which of the following is a×b\mathbf{a} \times \mathbf{b}?
Show worked solution

Worked solution

  1. Write down the two vectors

    a=(6−52),b=(−437)\mathbf{a}=\begin{pmatrix} 6 \\ -5 \\ 2 \end{pmatrix},\quad \mathbf{b}=\begin{pmatrix} -4 \\ 3 \\ 7 \end{pmatrix}

    Identify the components of each vector before expanding.

  2. Set up the determinant for the vector product

    a×b=∣ijk6−52−437∣\mathbf{a}\times \mathbf{b}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & -5 & 2 \\ -4 & 3 & 7 \end{vmatrix}

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

  3. Expand the i\mathbf{i} component

    (−5)(7)−(2)(3)=−41(-5)(7)-(2)(3)=-41

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

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

    −[(6)(7)−(2)(−4)]=−50-\big[(6)(7)-(2)(-4)\big]=-50

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

  5. Expand the k\mathbf{k} component

    (6)(3)−(−5)(−4)=−2(6)(3)-(-5)(-4)=-2

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

  6. Assemble the vector product

    a×b=(−41−50−2)\mathbf{a}\times \mathbf{b}=\begin{pmatrix} -41 \\ -50 \\ -2 \end{pmatrix}

    Collect the three components into a single column vector.

  7. Expand the k\mathbf{k} component

    (6)(3)−(−5)(−4)=−2(6)(3)-(-5)(-4)=-2

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

  8. Assemble the vector product

    a×b=(−41−50−2)\mathbf{a}\times \mathbf{b}=\begin{pmatrix} -41 \\ -50 \\ -2 \end{pmatrix}

    Collect the three components into a single column vector.

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

    a⋅(a×b)=(6)(−41)+(−5)(−50)+(2)(−2)=0\mathbf{a}\cdot(\mathbf{a}\times \mathbf{b})=(6)(-41)+(-5)(-50)+(2)(-2)=0

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

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

    b⋅(a×b)=(−4)(−41)+(3)(−50)+(7)(−2)=0\mathbf{b}\cdot(\mathbf{a}\times \mathbf{b})=(-4)(-41)+(3)(-50)+(7)(-2)=0

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

  11. Confirm the anticommutative property

    b×a=(41502)\mathbf{b}\times \mathbf{a}=\begin{pmatrix} 41 \\ 50 \\ 2 \end{pmatrix}

    Reversing the order of the vectors reverses every component.

  12. Find the magnitude of the vector product

    ∣a×b∣=−412+−502+−22=4185=3465\left|\mathbf{a}\times \mathbf{b}\right|=\sqrt{-41^2+-50^2+-2^2}=\sqrt{4185}=3\sqrt{465}

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

  13. Interpret the magnitude geometrically

    Area of parallelogram=3465\text{Area of parallelogram}=3\sqrt{465}

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

  14. Note the scalar product of the two vectors

    a⋅b=−25\mathbf{a}\cdot \mathbf{b}=-25

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

  15. Select the correct vector product

    (−41−50−2)\begin{pmatrix} -41 \\ -50 \\ -2 \end{pmatrix}

    This matches the expansion of the determinant.

Answer
(−41−50−2)\begin{pmatrix} -41 \\ -50 \\ -2 \end{pmatrix}

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