Hard A-Level Vectors in three dimensions Questions

Challenging, exam-style A-Level Vectors in three dimensions questions with worked solutions. Stretch yourself on the hardest vectors-3d, angle, distance, parallel problems.

vectors-3dangledistanceparallelmagnitudesolve
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
Find the distance between the points P=(1, 2, 2)P=\left(1,\ 2,\ 2\right) and Q=(7, 10, 11)Q=\left(7,\ 10,\ 11\right).
Show worked solution

Worked solution

  1. Find the displacement vector

    PQ→=q−p=6i+8j+9k\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}=6\mathbf{i}+8\mathbf{j}+9\mathbf{k}

    Subtract the position vectors of P and Q.

  2. Apply the distance formula

    PQ=∣PQ→∣=(6)2+(8)2+(9)2PQ=|\overrightarrow{PQ}|=\sqrt{\left(6\right)^{2}+\left(8\right)^{2}+\left(9\right)^{2}}

    The distance is the magnitude of the displacement vector.

  3. Write the position vector of the first point

    p=i+2j+2k\mathbf{p}=\mathbf{i}+2\mathbf{j}+2\mathbf{k}

    The position vector uses the point's coordinates as components.

  4. Write the position vector of the second point

    q=7i+10j+11k\mathbf{q}=7\mathbf{i}+10\mathbf{j}+11\mathbf{k}

    The position vector uses the point's coordinates as components.

  5. Find the displacement vector between the points

    PQ→=q−p=6i+8j+9k\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}=6\mathbf{i}+8\mathbf{j}+9\mathbf{k}

    Subtract the position vectors to get the displacement.

  6. Recall the distance formula

    PQ=(x2−x1)2+(y2−y1)2+(z2−z1)2PQ=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}+(z_2-z_1)^{2}}

    The distance is the magnitude of the displacement vector.

  7. Recall the midpoint formula

    M=(x1+x22, y1+y22, z1+z22)M=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2},\ \frac{z_1+z_2}{2}\right)

    The midpoint averages the coordinates.

  8. Write the components of the first vector

    a=6i+8j+9k\mathbf{a}=6\mathbf{i}+8\mathbf{j}+9\mathbf{k}

    Identify each i, j and k component.

  9. Recall the magnitude formula

    ∣v∣=v12+v22+v32|\mathbf{v}|=\sqrt{v_1^{2}+v_2^{2}+v_3^{2}}

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

  10. Square the components of the first vector

    (6)2+(8)2+(9)2=181\left(6\right)^{2}+\left(8\right)^{2}+\left(9\right)^{2}=181

    Squaring and adding gives the value under the root.

  11. Find the magnitude of the first vector

    ∣a∣=181=181|\mathbf{a}|=\sqrt{181}=\sqrt{181}

    Take the square root of the sum of squares.

  12. Find twice the first vector

    2a=12i+16j+18k2\mathbf{a}=12\mathbf{i}+16\mathbf{j}+18\mathbf{k}

    Multiply each component by 2.

  13. Find three times the first vector

    3a=18i+24j+27k3\mathbf{a}=18\mathbf{i}+24\mathbf{j}+27\mathbf{k}

    Multiply each component by 3.

  14. Find the negative of the first vector

    −a=−6i−8j−9k-\mathbf{a}=-6\mathbf{i}-8\mathbf{j}-9\mathbf{k}

    Reverse the sign of each component.

  15. State the distance

    PQ=181=181PQ=\sqrt{181}=\sqrt{181}

    This is the distance between the two points.

Answer
181\sqrt{181}
Question 2
8 markschallenging
Find the scalar product a⋅b\mathbf{a}\cdot\mathbf{b} where a=5i−2j+3k\mathbf{a}=5\mathbf{i}-2\mathbf{j}+3\mathbf{k} and b=i+4j−2k\mathbf{b}=\mathbf{i}+4\mathbf{j}-2\mathbf{k}.
Show worked solution

Worked solution

  1. Write the two vectors

    a=5i−2j+3k,b=i+4j−2k\mathbf{a}=5\mathbf{i}-2\mathbf{j}+3\mathbf{k},\quad \mathbf{b}=\mathbf{i}+4\mathbf{j}-2\mathbf{k}

    List both vectors in component form.

  2. Multiply corresponding components and add

    (5)(1)+(−2)(4)+(3)(−2)\left(5\right)\left(1\right)+\left(-2\right)\left(4\right)+\left(3\right)\left(-2\right)

    The scalar product multiplies matching components.

  3. Write the components of the first vector

    a=5i−2j+3k\mathbf{a}=5\mathbf{i}-2\mathbf{j}+3\mathbf{k}

    Identify each i, j and k component.

  4. Recall the magnitude formula

    ∣v∣=v12+v22+v32|\mathbf{v}|=\sqrt{v_1^{2}+v_2^{2}+v_3^{2}}

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

  5. Square the components of the first vector

    (5)2+(−2)2+(3)2=38\left(5\right)^{2}+\left(-2\right)^{2}+\left(3\right)^{2}=38

    Squaring and adding gives the value under the root.

  6. Find the magnitude of the first vector

    ∣a∣=38=38|\mathbf{a}|=\sqrt{38}=\sqrt{38}

    Take the square root of the sum of squares.

  7. Find twice the first vector

    2a=10i−4j+6k2\mathbf{a}=10\mathbf{i}-4\mathbf{j}+6\mathbf{k}

    Multiply each component by 2.

  8. Find three times the first vector

    3a=15i−6j+9k3\mathbf{a}=15\mathbf{i}-6\mathbf{j}+9\mathbf{k}

    Multiply each component by 3.

  9. Find the negative of the first vector

    −a=−5i+2j−3k-\mathbf{a}=-5\mathbf{i}+2\mathbf{j}-3\mathbf{k}

    Reverse the sign of each component.

  10. Recall the unit vector definition

    a^=a∣a∣\hat{\mathbf{a}}=\frac{\mathbf{a}}{|\mathbf{a}|}

    A unit vector points the same way but has magnitude one.

  11. Find the unit vector in the direction of a

    a^=138(5i−2j+3k)\hat{\mathbf{a}}=\frac{1}{\sqrt{38}}\left(5\mathbf{i}-2\mathbf{j}+3\mathbf{k}\right)

    Divide the vector by its magnitude.

  12. Write the first vector in column form

    (5−23)\begin{pmatrix} 5 \\ -2 \\ 3 \end{pmatrix}

    Column form lists the components vertically.

  13. Write the components of the second vector

    b=i+4j−2k\mathbf{b}=\mathbf{i}+4\mathbf{j}-2\mathbf{k}

    Identify each component of the second vector.

  14. Square the components of the second vector

    (1)2+(4)2+(−2)2=21\left(1\right)^{2}+\left(4\right)^{2}+\left(-2\right)^{2}=21

    Squaring and adding gives the value under the root.

  15. State the scalar product

    a⋅b=−9\mathbf{a}\cdot\mathbf{b}=-9

    This is the required scalar (dot) product.

Answer
−9-9
Question 3
8 markschallenging
Find the angle between a=2i+k\mathbf{a}=2\mathbf{i}+\mathbf{k} and b=i+j+k\mathbf{b}=\mathbf{i}+\mathbf{j}+\mathbf{k}, giving your answer in degrees to 1 decimal place.
Show worked solution

Worked solution

  1. Write the two vectors

    a=2i+k,b=i+j+k\mathbf{a}=2\mathbf{i}+\mathbf{k},\quad \mathbf{b}=\mathbf{i}+\mathbf{j}+\mathbf{k}

    List both vectors in component form.

  2. Find the scalar product

    a⋅b=3\mathbf{a}\cdot\mathbf{b}=3

    Multiply matching components and add.

  3. Find the magnitudes

    ∣a∣=5,∣b∣=3|\mathbf{a}|=\sqrt{5},\quad |\mathbf{b}|=\sqrt{3}

    The angle formula needs both magnitudes.

  4. Apply the angle formula

    cos⁡θ=315\cos\theta=\frac{3}{\sqrt{15}}

    Substitute the scalar product and magnitudes.

  5. Write the components of the first vector

    a=2i+k\mathbf{a}=2\mathbf{i}+\mathbf{k}

    Identify each i, j and k component.

  6. Recall the magnitude formula

    ∣v∣=v12+v22+v32|\mathbf{v}|=\sqrt{v_1^{2}+v_2^{2}+v_3^{2}}

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

  7. Square the components of the first vector

    (2)2+(0)2+(1)2=5\left(2\right)^{2}+\left(0\right)^{2}+\left(1\right)^{2}=5

    Squaring and adding gives the value under the root.

  8. Find the magnitude of the first vector

    ∣a∣=5=5|\mathbf{a}|=\sqrt{5}=\sqrt{5}

    Take the square root of the sum of squares.

  9. Find twice the first vector

    2a=4i+2k2\mathbf{a}=4\mathbf{i}+2\mathbf{k}

    Multiply each component by 2.

  10. Find three times the first vector

    3a=6i+3k3\mathbf{a}=6\mathbf{i}+3\mathbf{k}

    Multiply each component by 3.

  11. Find the negative of the first vector

    −a=−2i−k-\mathbf{a}=-2\mathbf{i}-\mathbf{k}

    Reverse the sign of each component.

  12. Recall the unit vector definition

    a^=a∣a∣\hat{\mathbf{a}}=\frac{\mathbf{a}}{|\mathbf{a}|}

    A unit vector points the same way but has magnitude one.

  13. Find the unit vector in the direction of a

    a^=15(2i+k)\hat{\mathbf{a}}=\frac{1}{\sqrt{5}}\left(2\mathbf{i}+\mathbf{k}\right)

    Divide the vector by its magnitude.

  14. Write the first vector in column form

    (201)\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}

    Column form lists the components vertically.

  15. State the angle to 1 decimal place

    θ=39.2∘\theta=39.2^{\circ}

    Take the inverse cosine and convert to degrees.

Answer
39.2∘39.2^{\circ}
Question 4
8 markschallenging
Which of the following equals a+b\mathbf{a}+\mathbf{b}, where a=i−2j+5k\mathbf{a}=\mathbf{i}-2\mathbf{j}+5\mathbf{k} and b=3i+2j−k\mathbf{b}=3\mathbf{i}+2\mathbf{j}-\mathbf{k}?
Show worked solution

Worked solution

  1. Write the two vectors

    a=i−2j+5k,b=3i+2j−k\mathbf{a}=\mathbf{i}-2\mathbf{j}+5\mathbf{k},\quad \mathbf{b}=3\mathbf{i}+2\mathbf{j}-\mathbf{k}

    List both vectors in component form.

  2. Add the corresponding components

    (1+3)i+(−2+2)j+(5−1)k=4i+4k\left(1+3\right)\mathbf{i}+\left(-2+2\right)\mathbf{j}+\left(5-1\right)\mathbf{k}=4\mathbf{i}+4\mathbf{k}

    Add the i, j and k components separately.

  3. Write the components of the first vector

    a=i−2j+5k\mathbf{a}=\mathbf{i}-2\mathbf{j}+5\mathbf{k}

    Identify each i, j and k component.

  4. Recall the magnitude formula

    ∣v∣=v12+v22+v32|\mathbf{v}|=\sqrt{v_1^{2}+v_2^{2}+v_3^{2}}

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

  5. Square the components of the first vector

    (1)2+(−2)2+(5)2=30\left(1\right)^{2}+\left(-2\right)^{2}+\left(5\right)^{2}=30

    Squaring and adding gives the value under the root.

  6. Find the magnitude of the first vector

    ∣a∣=30=30|\mathbf{a}|=\sqrt{30}=\sqrt{30}

    Take the square root of the sum of squares.

  7. Find twice the first vector

    2a=2i−4j+10k2\mathbf{a}=2\mathbf{i}-4\mathbf{j}+10\mathbf{k}

    Multiply each component by 2.

  8. Find three times the first vector

    3a=3i−6j+15k3\mathbf{a}=3\mathbf{i}-6\mathbf{j}+15\mathbf{k}

    Multiply each component by 3.

  9. Find the negative of the first vector

    −a=−i+2j−5k-\mathbf{a}=-\mathbf{i}+2\mathbf{j}-5\mathbf{k}

    Reverse the sign of each component.

  10. Recall the unit vector definition

    a^=a∣a∣\hat{\mathbf{a}}=\frac{\mathbf{a}}{|\mathbf{a}|}

    A unit vector points the same way but has magnitude one.

  11. Find the unit vector in the direction of a

    a^=130(i−2j+5k)\hat{\mathbf{a}}=\frac{1}{\sqrt{30}}\left(\mathbf{i}-2\mathbf{j}+5\mathbf{k}\right)

    Divide the vector by its magnitude.

  12. Write the first vector in column form

    (1−25)\begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix}

    Column form lists the components vertically.

  13. Write the components of the second vector

    b=3i+2j−k\mathbf{b}=3\mathbf{i}+2\mathbf{j}-\mathbf{k}

    Identify each component of the second vector.

  14. Square the components of the second vector

    (3)2+(2)2+(−1)2=14\left(3\right)^{2}+\left(2\right)^{2}+\left(-1\right)^{2}=14

    Squaring and adding gives the value under the root.

  15. Select the correct resultant

    a+b=4i+4k\mathbf{a}+\mathbf{b}=4\mathbf{i}+4\mathbf{k}

    This is the sum of the two vectors.

Answer
4i+4k4\mathbf{i}+4\mathbf{k}
Question 5
8 markschallenging
Which of the following equals a+b\mathbf{a}+\mathbf{b}, where a=2i+3j+k\mathbf{a}=2\mathbf{i}+3\mathbf{j}+\mathbf{k} and b=4i−j+2k\mathbf{b}=4\mathbf{i}-\mathbf{j}+2\mathbf{k}?
Show worked solution

Worked solution

  1. Write the two vectors

    a=2i+3j+k,b=4i−j+2k\mathbf{a}=2\mathbf{i}+3\mathbf{j}+\mathbf{k},\quad \mathbf{b}=4\mathbf{i}-\mathbf{j}+2\mathbf{k}

    List both vectors in component form.

  2. Add the corresponding components

    (2+4)i+(3−1)j+(1+2)k=6i+2j+3k\left(2+4\right)\mathbf{i}+\left(3-1\right)\mathbf{j}+\left(1+2\right)\mathbf{k}=6\mathbf{i}+2\mathbf{j}+3\mathbf{k}

    Add the i, j and k components separately.

  3. Write the components of the first vector

    a=2i+3j+k\mathbf{a}=2\mathbf{i}+3\mathbf{j}+\mathbf{k}

    Identify each i, j and k component.

  4. Recall the magnitude formula

    ∣v∣=v12+v22+v32|\mathbf{v}|=\sqrt{v_1^{2}+v_2^{2}+v_3^{2}}

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

  5. Square the components of the first vector

    (2)2+(3)2+(1)2=14\left(2\right)^{2}+\left(3\right)^{2}+\left(1\right)^{2}=14

    Squaring and adding gives the value under the root.

  6. Find the magnitude of the first vector

    ∣a∣=14=14|\mathbf{a}|=\sqrt{14}=\sqrt{14}

    Take the square root of the sum of squares.

  7. Find twice the first vector

    2a=4i+6j+2k2\mathbf{a}=4\mathbf{i}+6\mathbf{j}+2\mathbf{k}

    Multiply each component by 2.

  8. Find three times the first vector

    3a=6i+9j+3k3\mathbf{a}=6\mathbf{i}+9\mathbf{j}+3\mathbf{k}

    Multiply each component by 3.

  9. Find the negative of the first vector

    −a=−2i−3j−k-\mathbf{a}=-2\mathbf{i}-3\mathbf{j}-\mathbf{k}

    Reverse the sign of each component.

  10. Recall the unit vector definition

    a^=a∣a∣\hat{\mathbf{a}}=\frac{\mathbf{a}}{|\mathbf{a}|}

    A unit vector points the same way but has magnitude one.

  11. Find the unit vector in the direction of a

    a^=114(2i+3j+k)\hat{\mathbf{a}}=\frac{1}{\sqrt{14}}\left(2\mathbf{i}+3\mathbf{j}+\mathbf{k}\right)

    Divide the vector by its magnitude.

  12. Write the first vector in column form

    (231)\begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}

    Column form lists the components vertically.

  13. Write the components of the second vector

    b=4i−j+2k\mathbf{b}=4\mathbf{i}-\mathbf{j}+2\mathbf{k}

    Identify each component of the second vector.

  14. Square the components of the second vector

    (4)2+(−1)2+(2)2=21\left(4\right)^{2}+\left(-1\right)^{2}+\left(2\right)^{2}=21

    Squaring and adding gives the value under the root.

  15. Select the correct resultant

    a+b=6i+2j+3k\mathbf{a}+\mathbf{b}=6\mathbf{i}+2\mathbf{j}+3\mathbf{k}

    This is the sum of the two vectors.

Answer
6i+2j+3k6\mathbf{i}+2\mathbf{j}+3\mathbf{k}

Unlock 29 more Vectors in three dimensions questions

Create a free account to work through every A-Level Vectors in three dimensions question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Vectors in three dimensions practice

Practice QuestionsWorked Solutions

Related Pure Maths topics

Areas and further applicationsMathematical argument and notationArithmetic sequences and seriesBinomial expansion (rational powers)Binomial expansionCircles