Vectors: lines and planes Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Vectors: lines and planes questions. See exactly how to solve problems on scalar-product, vectors, vector-equation, line-through-two-points.

scalar-productvectorsvector-equationline-through-two-pointspoint-on-lineplane-equation
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Given a=(23−1)\mathbf{a}=\begin{pmatrix}2\\3\\-1\end{pmatrix} and b=(415)\mathbf{b}=\begin{pmatrix}4\\1\\5\end{pmatrix}, find the scalar product a⋅b\mathbf{a}\cdot\mathbf{b}.

Worked solution

  1. Write down the two vectors

    a=(23−1),b=(415)\mathbf{a}=\begin{pmatrix}2\\3\\-1\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}4\\1\\5\end{pmatrix}

    Identify the components of each vector.

  2. Multiply corresponding components

    (2)(4), (3)(1), (−1)(5)(2)(4),\ (3)(1),\ (-1)(5)

    The scalar product pairs up matching components.

  3. Add the three products

    a⋅b=8+3−5=6\mathbf{a}\cdot\mathbf{b}=8+3-5=6

    Summing the products gives the scalar product.

  4. State the scalar product

    a⋅b=6\mathbf{a}\cdot\mathbf{b}=6

    This is the required value of the scalar product.

Answer
66
Question 2
2 markseasy
Given a=(1−23)\mathbf{a}=\begin{pmatrix}1\\-2\\3\end{pmatrix} and b=(254)\mathbf{b}=\begin{pmatrix}2\\5\\4\end{pmatrix}, find the scalar product a⋅b\mathbf{a}\cdot\mathbf{b}.

Worked solution

  1. Write down the two vectors

    a=(1−23),b=(254)\mathbf{a}=\begin{pmatrix}1\\-2\\3\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}2\\5\\4\end{pmatrix}

    Identify the components of each vector.

  2. Multiply corresponding components

    (1)(2), (−2)(5), (3)(4)(1)(2),\ (-2)(5),\ (3)(4)

    The scalar product pairs up matching components.

  3. Add the three products

    a⋅b=2−10+12=4\mathbf{a}\cdot\mathbf{b}=2-10+12=4

    Summing the products gives the scalar product.

  4. State the scalar product

    a⋅b=4\mathbf{a}\cdot\mathbf{b}=4

    This is the required value of the scalar product.

Answer
44
Question 3
2 markseasy
Find a vector equation of the line that passes through the points A(1,2,3)A(1,2,3) and B(3,5,7)B(3,5,7).

Worked solution

  1. Write the position vectors of the two points

    a=(123),b=(357)\mathbf{a}=\begin{pmatrix}1\\2\\3\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}3\\5\\7\end{pmatrix}

    The coordinates of AA and BB become position vectors.

  2. Find the direction vector AB→\overrightarrow{AB}

    AB→=b−a=(234)\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}2\\3\\4\end{pmatrix}

    Subtract the position vector of AA from that of BB.

  3. Simplify the direction vector

    d=(234)\mathbf{d}=\begin{pmatrix}2\\3\\4\end{pmatrix}

    Any non-zero multiple of AB→\overrightarrow{AB} may be used as the direction.

  4. State a vector equation of the line

    r=(123)+λ(234)\mathbf{r}=\begin{pmatrix}1\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}2\\3\\4\end{pmatrix}

    Any point of the line together with its direction gives a valid equation.

Answer
r=(123)+λ(234)\mathbf{r}=\begin{pmatrix}1\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}2\\3\\4\end{pmatrix}
Question 4
2 markseasy
Find a vector equation of the line that passes through the points A(2,−1,4)A(2,-1,4) and B(6,1,0)B(6,1,0).

Worked solution

  1. Write the position vectors of the two points

    a=(2−14),b=(610)\mathbf{a}=\begin{pmatrix}2\\-1\\4\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}6\\1\\0\end{pmatrix}

    The coordinates of AA and BB become position vectors.

  2. Find the direction vector AB→\overrightarrow{AB}

    AB→=b−a=(42−4)\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}4\\2\\-4\end{pmatrix}

    Subtract the position vector of AA from that of BB.

  3. Simplify the direction vector

    d=(21−2)\mathbf{d}=\begin{pmatrix}2\\1\\-2\end{pmatrix}

    Any non-zero multiple of AB→\overrightarrow{AB} may be used as the direction.

  4. State a vector equation of the line

    r=(2−14)+λ(21−2)\mathbf{r}=\begin{pmatrix}2\\-1\\4\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-2\end{pmatrix}

    Any point of the line together with its direction gives a valid equation.

Answer
r=(2−14)+λ(21−2)\mathbf{r}=\begin{pmatrix}2\\-1\\4\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-2\end{pmatrix}
Question 5
2 markseasy
The line ll has vector equation r=(102)+λ(2−13)\mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\3\end{pmatrix}. Find the position vector of the point on ll where λ=3\lambda=3.

Worked solution

  1. Write down the line equation

    r=(102)+λ(2−13)\mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\3\end{pmatrix}

    Identify the base point and direction vector.

  2. Substitute the given value of the parameter

    r=(102)+(3)(2−13)\mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+(3)\begin{pmatrix}2\\-1\\3\end{pmatrix}

    Replace λ\lambda by the value given in the question.

  3. Multiply the direction vector by the parameter

    (3)(2−13)=(6−39)(3)\begin{pmatrix}2\\-1\\3\end{pmatrix}=\begin{pmatrix}6\\-3\\9\end{pmatrix}

    Scalar multiplication multiplies each component.

  4. State the position vector

    r=(7−311)\mathbf{r}=\begin{pmatrix}7\\-3\\11\end{pmatrix}

    This is the position vector of the required point.

Answer
(7−311)\begin{pmatrix}7\\-3\\11\end{pmatrix}

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