A-Level Vectors in three dimensions Practice Questions

Free A-Level Vectors in three dimensions practice questions with full step-by-step worked solutions. Covers vectors-3d, resultant, magnitude, dot-product. Practise exam-style problems and check your method.

vectors-3dresultantmagnitudedot-productdistancemidpoint
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Given a=2i+j+3k\mathbf{a}=2\mathbf{i}+\mathbf{j}+3\mathbf{k} and b=i−2j+4k\mathbf{b}=\mathbf{i}-2\mathbf{j}+4\mathbf{k}, find a+b\mathbf{a}+\mathbf{b}.
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Worked solution

  1. Write down the given vectors

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

    List the two vectors in component form.

  2. Combine the corresponding components

    (2+1)i+(1−2)j+(3+4)k=3i−j+7k\left(2+1\right)\mathbf{i}+\left(1-2\right)\mathbf{j}+\left(3+4\right)\mathbf{k}=3\mathbf{i}-\mathbf{j}+7\mathbf{k}

    Group the i, j and k components separately.

  3. State the resultant vector

    a+b=3i−j+7k\mathbf{a}+\mathbf{b}=3\mathbf{i}-\mathbf{j}+7\mathbf{k}

    This is the required resultant vector.

Answer
3i−j+7k3\mathbf{i}-\mathbf{j}+7\mathbf{k}
Question 2
2 markseasy
Which of the following equals a+b\mathbf{a}+\mathbf{b}, where a=i+2j−k\mathbf{a}=\mathbf{i}+2\mathbf{j}-\mathbf{k} and b=3i+2k\mathbf{b}=3\mathbf{i}+2\mathbf{k}?
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Worked solution

  1. Write the two vectors

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

    List both vectors in component form.

  2. Add the corresponding components

    (1+3)i+(2+0)j+(−1+2)k=4i+2j+k\left(1+3\right)\mathbf{i}+\left(2+0\right)\mathbf{j}+\left(-1+2\right)\mathbf{k}=4\mathbf{i}+2\mathbf{j}+\mathbf{k}

    Add the i, j and k components separately.

  3. Select the correct resultant

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

    This is the sum of the two vectors.

Answer
4i+2j+k4\mathbf{i}+2\mathbf{j}+\mathbf{k}
Question 3
3 marksintermediate
Which of the following is a unit vector?
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Worked solution

  1. Recall that a unit vector has magnitude one

    ∣a^∣=1|\hat{\mathbf{a}}|=1

    A unit vector has length exactly one.

  2. Check the magnitude of the correct option

    1=1\sqrt{1}=1

    Its components give a sum of squares equal to 1.

  3. Write the components of the first vector

    a=13i+23j+23k\mathbf{a}=\frac{1}{3}\mathbf{i}+\frac{2}{3}\mathbf{j}+\frac{2}{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

    (13)2+(23)2+(23)2=1\left(\frac{1}{3}\right)^{2}+\left(\frac{2}{3}\right)^{2}+\left(\frac{2}{3}\right)^{2}=1

    Squaring and adding gives the value under the root.

  6. Select the unit vector

    13i+23j+23k\frac{1}{3}\mathbf{i}+\frac{2}{3}\mathbf{j}+\frac{2}{3}\mathbf{k}

    Only this option has magnitude one.

Answer
13i+23j+23k\frac{1}{3}\mathbf{i}+\frac{2}{3}\mathbf{j}+\frac{2}{3}\mathbf{k}
Question 4
5 markshard
Find the coordinates of the midpoint of A=(1, 2, 3)A=\left(1,\ 2,\ 3\right) and B=(5, 8, 11)B=\left(5,\ 8,\ 11\right).
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Worked solution

  1. Write the coordinates of the two points

    P=(1, 2, 3),Q=(5, 8, 11)P=\left(1,\ 2,\ 3\right),\quad Q=\left(5,\ 8,\ 11\right)

    Note the coordinates of each point.

  2. Apply the midpoint formula

    M=(1+52, 2+82, 3+112)M=\left(\frac{1+5}{2},\ \frac{2+8}{2},\ \frac{3+11}{2}\right)

    Average the coordinates of the two points.

  3. Write the position vector of the first point

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

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

  4. Write the position vector of the second point

    q=5i+8j+11k\mathbf{q}=5\mathbf{i}+8\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=4i+6j+8k\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}=4\mathbf{i}+6\mathbf{j}+8\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=4i+6j+8k\mathbf{a}=4\mathbf{i}+6\mathbf{j}+8\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. State the midpoint

    M=(3, 5, 7)M=\left(3,\ 5,\ 7\right)

    This is the midpoint of the segment.

Answer
(3, 5, 7)\left(3,\ 5,\ 7\right)
Question 5
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).
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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}

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