Hard Further Maths Vectors: lines and planes Questions

Challenging, exam-style Further Maths Vectors: lines and planes questions with worked solutions. Stretch yourself on the hardest intersection, two-lines, distance, point-to-line problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
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)
Question 2
8 markschallenging
The planes Π1\Pi_1 and Π2\Pi_2 have equations x−2y+2z=5x-2y+2z=5 and 3x+4y−12z=13x+4y-12z=1. Which of the following is the acute angle between Π1\Pi_1 and Π2\Pi_2, in degrees to 1 decimal place?
Show worked solution

Worked solution

  1. Read the normal vector of each plane

    n1=(1−22),n2=(34−12)\mathbf{n}_1=\begin{pmatrix}1\\-2\\2\end{pmatrix},\quad \mathbf{n}_2=\begin{pmatrix}3\\4\\-12\end{pmatrix}

    The plane coefficients give the normals.

  2. Evaluate the scalar product of the normals

    n1⋅n2=−29\mathbf{n}_1\cdot\mathbf{n}_2=-29

    Multiply matching components and add.

  3. Find the magnitudes of the normals

    ∣n1∣=3,∣n2∣=13\left|\mathbf{n}_1\right|=3,\quad \left|\mathbf{n}_2\right|=13

    Apply the magnitude formula.

  4. Substitute into the cosine formula

    cos⁡θ=293×13=0.743590\cos\theta=\frac{29}{3\times13}=0.743590

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

  5. Take the inverse cosine

    θ=cos⁡−1(0.743590)=42.0∘\theta=\cos^{-1}\left(0.743590\right)=42.0^{\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. 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.

  13. 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.

  14. 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.

  15. 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.

  16. Select the acute angle between the planes

    42.0∘42.0^{\circ}

    This is the angle found from the normals.

Answer
42.0∘42.0^{\circ}
Question 3
8 markschallenging
Which of the following is the exact perpendicular distance from the point P(−2,5,1)P(-2,5,1) to the plane x−3y+2z=4x-3y+2z=4?
Show worked solution

Worked solution

  1. Write down the normal vector and the constant

    n=(1−32),d=4\mathbf{n}=\begin{pmatrix}1\\-3\\2\end{pmatrix},\quad d=4

    Read them from the Cartesian equation of the plane.

  2. Use the point-to-plane distance formula

    D=∣n⋅p−d∣∣n∣D=\frac{\left|\mathbf{n}\cdot\mathbf{p}-d\right|}{\left|\mathbf{n}\right|}

    This measures the perpendicular distance from PP to the plane.

  3. Evaluate the numerator

    ∣−15−(4)∣=19\left|-15-(4)\right|=19

    Substitute the coordinates of PP into the plane expression.

  4. Evaluate the denominator

    ∣n∣=14\left|\mathbf{n}\right|=\sqrt{14}

    Apply the magnitude formula to the normal.

  5. Divide and simplify the surd

    D=1914=191414D=\frac{19}{\sqrt{14}}=\frac{19\sqrt{14}}{14}

    Rationalise the denominator to get the exact value.

  6. Reject options that forget the denominator

    ∣n⋅p−d∣=19\left|\mathbf{n}\cdot\mathbf{p}-d\right|=19

    The numerator alone is not a distance.

  7. Give a decimal check

    D≈5.078D\approx5.078

    A decimal value confirms which option has the right size.

  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. 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.

  13. 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.

  14. 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.

  15. 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.

  16. Select the exact perpendicular distance

    191414\frac{19\sqrt{14}}{14}

    This is the value given by the point-to-plane distance formula.

Answer
191414\frac{19\sqrt{14}}{14}
Question 4
8 markschallenging
The line ll has equation r=(12−1)+λ(312)\mathbf{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}3\\1\\2\end{pmatrix} and the plane Π\Pi has equation 2x−y+z=42x-y+z=4. Which one of the following statements about ll and Π\Pi is correct?
Show worked solution

Worked solution

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

    d=(312),n=(2−11)\mathbf{d}=\begin{pmatrix}3\\1\\2\end{pmatrix},\quad \mathbf{n}=\begin{pmatrix}2\\-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=7\mathbf{d}\cdot\mathbf{n}=7

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

  3. Test whether the base point lies in the plane

    2(1)−1(2)+1(−1)=−12\left(1\right)-1\left(2\right)+1\left(-1\right)=-1

    Compare this value with the constant 44 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. Check whether the direction is parallel to the normal

    d≠tn\mathbf{d}\neq t\mathbf{n}

    That would make the line perpendicular to the plane.

  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. 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.

  13. 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.

  14. 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.

  15. 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.

  16. Select the correct statement

    The line meets the plane at exactly one point\text{The line meets the plane at exactly one point}

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

Answer
The line meets the plane at exactly one point\text{The line meets the plane at exactly one point}
Question 5
8 markschallenging
The lines l1l_1 and l2l_2 have equations r=(1−12)+λ(231)\mathbf{r}=\begin{pmatrix}1\\-1\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\3\\1\end{pmatrix} and r=(416)+μ(1−13)\mathbf{r}=\begin{pmatrix}4\\1\\6\end{pmatrix}+\mu\begin{pmatrix}1\\-1\\3\end{pmatrix} respectively. Which one of the following statements about l1l_1 and l2l_2 is correct?
Show worked solution

Worked solution

  1. Write the direction vectors

    d1=(231),d2=(1−13)\mathbf{d}_1=\begin{pmatrix}2\\3\\1\end{pmatrix},\quad \mathbf{d}_2=\begin{pmatrix}1\\-1\\3\end{pmatrix}

    Read the directions from the two vector equations.

  2. Test whether the lines are parallel

    d2≠td1\mathbf{d}_2\neq t\mathbf{d}_1

    Parallel lines have proportional direction vectors.

  3. Set the general points equal

    (1−12)+λ(231)=(416)+μ(1−13)\begin{pmatrix}1\\-1\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\3\\1\end{pmatrix}=\begin{pmatrix}4\\1\\6\end{pmatrix}+\mu\begin{pmatrix}1\\-1\\3\end{pmatrix}

    If the lines meet, some pair (λ,μ)(\lambda,\mu) satisfies all three components.

  4. Solve two of the component equations

    1+2λ=4+μ,−1+3λ=1−μ1+2\lambda=4+\mu,\quad -1+3\lambda=1-\mu

    Two equations determine λ\lambda and μ\mu when the lines are not parallel.

  5. Test the remaining component for consistency

    2+λ versus 6+3μ2+\lambda\ \text{versus}\ 6+3\mu

    If the third equation fails the lines cannot meet.

  6. Recall what skew means

    skew  ⟺  not parallel and no common point\text{skew}\iff\text{not parallel and no common point}

    Skew lines exist only in three or more dimensions.

  7. Check the scalar product of the directions

    d1⋅d2=2\mathbf{d}_1\cdot\mathbf{d}_2=2

    A zero value would mean the directions are perpendicular.

  8. Rule out coincident lines

    A1A2→=(324)\overrightarrow{A_1A_2}=\begin{pmatrix}3\\2\\4\end{pmatrix}

    Coincident lines would need this displacement to be along the common direction.

  9. 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.

  10. 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.

  11. 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.

  12. 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.

  13. 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.

  14. 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.

  15. 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.

  16. Select the correct statement

    The lines intersect at exactly one point\text{The lines intersect at exactly one point}

    This follows from the directions and the consistency of the equations.

Answer
The lines intersect at exactly one point\text{The lines intersect at exactly one point}

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