Further Maths Vectors: lines and planes Practice Questions

Free Further Maths Vectors: lines and planes practice questions with full step-by-step worked solutions. Covers scalar-product, vectors, vector-equation, line-through-two-points. Practise exam-style problems and check your method.

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}.
Show worked solution

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
The planes Π1\Pi_1 and Π2\Pi_2 have equations 2x+y−z=42x+y-z=4 and x−2y=3x-2y=3. Which one of the following statements about Π1\Pi_1 and Π2\Pi_2 is correct?
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Worked solution

  1. Read the normal vector of each plane

    n1=(21−1),n2=(1−20)\mathbf{n}_1=\begin{pmatrix}2\\1\\-1\end{pmatrix},\quad \mathbf{n}_2=\begin{pmatrix}1\\-2\\0\end{pmatrix}

    The coefficients of xx, yy and zz give the normals.

  2. Test whether the normals are parallel

    n2≠tn1\mathbf{n}_2\neq t\mathbf{n}_1

    Parallel normals mean parallel planes.

  3. Test whether the normals are perpendicular

    n1⋅n2=0\mathbf{n}_1\cdot\mathbf{n}_2=0

    A zero scalar product means the planes are perpendicular.

  4. Select the correct statement

    The planes are perpendicular\text{The planes are perpendicular}

    This follows from the normal vectors of the two planes.

Answer
The planes are perpendicular\text{The planes are perpendicular}
Question 3
4 marksintermediate
The line ll has equation r=(111)+λ(2−11)\mathbf{r}=\begin{pmatrix}1\\1\\1\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\1\end{pmatrix} and the plane Π\Pi has equation x+y−z=5x+y-z=5. Which one of the following statements about ll and Π\Pi is correct?
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Worked solution

  1. Write the direction of the line and the normal of the plane

    d=(2−11),n=(11−1)\mathbf{d}=\begin{pmatrix}2\\-1\\1\end{pmatrix},\quad \mathbf{n}=\begin{pmatrix}1\\1\\-1\end{pmatrix}

    Read the direction from the line and the normal from the plane.

  2. Test whether the direction is perpendicular to the normal

    d⋅n=0\mathbf{d}\cdot\mathbf{n}=0

    A zero value means the line is parallel to the plane.

  3. Test whether the base point lies in the plane

    1(1)+1(1)−1(1)=11\left(1\right)+1\left(1\right)-1\left(1\right)=1

    Compare this value with the constant 55 in the plane equation.

  4. Recall the condition for a line to lie in a plane

    d⋅n=0 and n⋅a=d\mathbf{d}\cdot\mathbf{n}=0\ \text{and}\ \mathbf{n}\cdot\mathbf{a}=d

    The direction must be parallel to the plane and the base point must lie in it.

  5. Recall the condition for a unique intersection

    d⋅n≠0\mathbf{d}\cdot\mathbf{n}\neq 0

    A non-zero scalar product forces exactly one point of intersection.

  6. Note a straight line cannot meet a plane twice unless it lies in it

    two common points⇒the line lies in the plane\text{two common points}\Rightarrow\text{the line lies in the plane}

    Two distinct common points would force the whole line into the plane.

  7. Select the correct statement

    The line is parallel to the plane and does not meet it\text{The line is parallel to the plane and does not meet it}

    This follows from the scalar product of the direction with the normal.

Answer
The line is parallel to the plane and does not meet it\text{The line is parallel to the plane and does not meet it}
Question 4
6 markshard
The planes Π1\Pi_1 and Π2\Pi_2 have equations 2x−y+2z=32x-y+2z=3 and x+2y+2z=1x+2y+2z=1. Which of the following is the acute angle between Π1\Pi_1 and Π2\Pi_2, in degrees to 1 decimal place?
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Worked solution

  1. Read the normal vector of each plane

    n1=(2−12),n2=(122)\mathbf{n}_1=\begin{pmatrix}2\\-1\\2\end{pmatrix},\quad \mathbf{n}_2=\begin{pmatrix}1\\2\\2\end{pmatrix}

    The plane coefficients give the normals.

  2. Evaluate the scalar product of the normals

    n1⋅n2=4\mathbf{n}_1\cdot\mathbf{n}_2=4

    Multiply matching components and add.

  3. Find the magnitudes of the normals

    ∣n1∣=3,∣n2∣=3\left|\mathbf{n}_1\right|=3,\quad \left|\mathbf{n}_2\right|=3

    Apply the magnitude formula.

  4. Substitute into the cosine formula

    cos⁡θ=43×3=0.444444\cos\theta=\frac{4}{3\times3}=0.444444

    Use the modulus of the scalar product to obtain the acute angle.

  5. Take the inverse cosine

    θ=cos⁡−1(0.444444)=63.6∘\theta=\cos^{-1}\left(0.444444\right)=63.6^{\circ}

    This gives the acute angle between the planes.

  6. Reject the complementary angle

    90∘−θ90^{\circ}-\theta

    The complement is the angle a line along one normal makes with the other plane.

  7. Reject the obtuse supplement

    180∘−θ180^{\circ}-\theta

    Convention requires the acute angle between two planes.

  8. Recall the scalar product in component form

    a⋅b=a1b1+a2b2+a3b3\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3

    The scalar product multiplies matching components and adds the results.

  9. Recall the magnitude of a vector

    ∣a∣=a12+a22+a32\left|\mathbf{a}\right|=\sqrt{a_1^2+a_2^2+a_3^2}

    The magnitude is the square root of the sum of the squares of the components.

  10. Recall the angle formula

    cos⁡θ=a⋅b∣a∣∣b∣\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{\left|\mathbf{a}\right|\left|\mathbf{b}\right|}

    The scalar product links the angle between two vectors to their magnitudes.

  11. Recall the perpendicularity test

    a⋅b=0  ⟺  a⊥b\mathbf{a}\cdot\mathbf{b}=0\iff\mathbf{a}\perp\mathbf{b}

    Two non-zero vectors are perpendicular exactly when their scalar product is zero.

  12. Select the acute angle between the planes

    63.6∘63.6^{\circ}

    This is the angle found from the normals.

Answer
63.6∘63.6^{\circ}
Question 5
8 markschallenging
Which of the following points lies in the plane 3x−2y+5z=73x-2y+5z=7?
Show worked solution

Worked solution

  1. Write the plane equation

    3x−2y+5z=73x-2y+5z=7

    A point lies in the plane only if its coordinates satisfy this equation.

  2. Substitute the coordinates of each option

    3x−2y+5z3x-2y+5z

    Evaluate the left-hand side for each candidate point.

  3. Identify the option giving the correct constant

    3(1)−2(3)+5(2)=73\left(1\right)-2\left(3\right)+5\left(2\right)=7

    This option balances the equation exactly.

  4. Reject the options that do not balance

    3x−2y+5z≠73x-2y+5z\neq7

    Any point giving a different value does not lie in the plane.

  5. Note the normal vector plays no part in this test

    n=(3−25)\mathbf{n}=\begin{pmatrix}3\\-2\\5\end{pmatrix}

    Membership of a plane is decided purely by substitution.

  6. Recall the scalar product in component form

    a⋅b=a1b1+a2b2+a3b3\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3

    The scalar product multiplies matching components and adds the results.

  7. Recall the magnitude of a vector

    ∣a∣=a12+a22+a32\left|\mathbf{a}\right|=\sqrt{a_1^2+a_2^2+a_3^2}

    The magnitude is the square root of the sum of the squares of the components.

  8. Recall the angle formula

    cos⁡θ=a⋅b∣a∣∣b∣\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{\left|\mathbf{a}\right|\left|\mathbf{b}\right|}

    The scalar product links the angle between two vectors to their magnitudes.

  9. Recall the perpendicularity test

    a⋅b=0  ⟺  a⊥b\mathbf{a}\cdot\mathbf{b}=0\iff\mathbf{a}\perp\mathbf{b}

    Two non-zero vectors are perpendicular exactly when their scalar product is zero.

  10. Recall the parallelism test

    d2=td1  ⟺  l1∥l2\mathbf{d}_2=t\mathbf{d}_1\iff l_1\parallel l_2

    Two lines are parallel exactly when their direction vectors are scalar multiples.

  11. Recall the vector equation of a line

    r=a+λd\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}

    A point on the line plus a multiple of the direction vector traces the whole line.

  12. Recall the Cartesian form of a line

    x−a1d1=y−a2d2=z−a3d3\frac{x-a_1}{d_1}=\frac{y-a_2}{d_2}=\frac{z-a_3}{d_3}

    Eliminating the parameter gives three equal expressions.

  13. Recall the scalar-product form of a plane

    r⋅n=d\mathbf{r}\cdot\mathbf{n}=d

    Every point of the plane has the same scalar product with the normal vector.

  14. Recall the Cartesian form of a plane

    ax+by+cz=dax+by+cz=d

    The coefficients of xx, yy and zz are the components of a normal vector.

  15. Read the normal vector from the Cartesian equation

    n=(abc)\mathbf{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}

    The normal is built directly from the coefficients in ax+by+cz=dax+by+cz=d.

  16. Select the point that lies in the plane

    (1,3,2)(1,3,2)

    Only this point satisfies the plane equation.

Answer
(1,3,2)(1,3,2)

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