Vector and triple products Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Vector and triple products questions. See exactly how to solve problems on vector-product, cross-product, anticommutativity, magnitude.

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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}.

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
Given a=(2−13)\mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} and b=(14−2)\mathbf{b} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}, find a×b\mathbf{a} \times \mathbf{b}.

Worked solution

  1. Set up the determinant for the vector product

    a×b=∣ijk2−1314−2∣\mathbf{a}\times \mathbf{b}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 1 & 4 & -2 \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

    (−1)(−2)−(3)(4)=−10(-1)(-2)-(3)(4)=-10

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

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

    −[(2)(−2)−(3)(1)]=7-\big[(2)(-2)-(3)(1)\big]=7

    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=(−1079)\mathbf{a}\times\mathbf{b}=\begin{pmatrix} -10 \\ 7 \\ 9 \end{pmatrix}

    This vector is perpendicular to both of the given vectors.

Answer
(−1079)\begin{pmatrix} -10 \\ 7 \\ 9 \end{pmatrix}
Question 3
2 markseasy
Given a=(301)\mathbf{a} = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} and b=(−254)\mathbf{b} = \begin{pmatrix} -2 \\ 5 \\ 4 \end{pmatrix}, find a×b\mathbf{a} \times \mathbf{b}.

Worked solution

  1. Set up the determinant for the vector product

    a×b=∣ijk301−254∣\mathbf{a}\times \mathbf{b}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & 1 \\ -2 & 5 & 4 \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

    (0)(4)−(1)(5)=−5(0)(4)-(1)(5)=-5

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

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

    −[(3)(4)−(1)(−2)]=−14-\big[(3)(4)-(1)(-2)\big]=-14

    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=(−5−1415)\mathbf{a}\times\mathbf{b}=\begin{pmatrix} -5 \\ -14 \\ 15 \end{pmatrix}

    This vector is perpendicular to both of the given vectors.

Answer
(−5−1415)\begin{pmatrix} -5 \\ -14 \\ 15 \end{pmatrix}
Question 4
2 markseasy
Given a=(112)\mathbf{a} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} and b=(3−10)\mathbf{b} = \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix}, find b×a\mathbf{b} \times \mathbf{a}.

Worked solution

  1. Set up the determinant for the vector product

    b×a=∣ijk3−10112∣\mathbf{b}\times \mathbf{a}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 0 \\ 1 & 1 & 2 \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

    (−1)(2)−(0)(1)=−2(-1)(2)-(0)(1)=-2

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

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

    −[(3)(2)−(0)(1)]=−6-\big[(3)(2)-(0)(1)\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

    b×a=(−2−64)\mathbf{b}\times\mathbf{a}=\begin{pmatrix} -2 \\ -6 \\ 4 \end{pmatrix}

    Note this is the negative of a×b\mathbf{a}\times\mathbf{b}.

Answer
(−2−64)\begin{pmatrix} -2 \\ -6 \\ 4 \end{pmatrix}
Question 5
2 markseasy
Given a=(10−1)\mathbf{a} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} and b=(231)\mathbf{b} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}, find the exact value of ∣a×b∣\left|\mathbf{a} \times \mathbf{b}\right|.

Worked solution

  1. Form the vector product

    a×b=∣ijk10−1231∣=(3−33)\mathbf{a}\times \mathbf{b}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -1 \\ 2 & 3 & 1 \end{vmatrix}=\begin{pmatrix} 3 \\ -3 \\ 3 \end{pmatrix}

    Expand the determinant to find the vector normal to both.

  2. Square each component

    32+−32+32=273^2+-3^2+3^2=27

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

  3. Take the square root

    ∣a×b∣=27=33\left|\mathbf{a}\times \mathbf{b}\right|=\sqrt{27}=3\sqrt{3}

    Simplify the surd by extracting any square factors.

  4. State the exact magnitude

    ∣a×b∣=33\left|\mathbf{a}\times\mathbf{b}\right|=3\sqrt{3}

    This is the exact value of the magnitude of the vector product.

Answer
333\sqrt{3}

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