P10.2Magnitude and direction

Edexcel A level Maths (9MA0) · Pure mathematics › Vectors

Practise Magnitude and direction. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy5 marks

A student finds the magnitude of the vector \(\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}\) as follows:
\(|\mathbf{a}| = \sqrt{2^2 - 3^2 + 6^2} = \sqrt{31}\)

(a) Explain the student's error and find the correct value of \(|\mathbf{a}|\).[2]

(b) Find a unit vector in the direction of \(\mathbf{a}\).[1]

Show the answer and mark scheme

(a) Answer: \((-3)^2 = 9\), so the squares must be added: \(|\mathbf{a}| = \sqrt{4 + 9 + 36} = 7\).

B1 for the square of \(-3\) is \(+9\): the magnitude is the square root of the sum of the squares of the components
B1 for 7

Worked solution: The \(\mathbf{j}\)-component is \(-3\), and \((-3)^2 = 9\), not \(-9\). \(|\mathbf{a}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\).

(b) Answer: \(\frac{1}{7}(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k})\)

B1 for \(\frac{2}{7}\mathbf{i} - \frac{3}{7}\mathbf{j} + \frac{6}{7}\mathbf{k}\) (ft their \(|\mathbf{a}|\))

Worked solution: \(\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|} = \dfrac{1}{7}(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k})\).

M1 for \(\cos\theta = \frac{2}{7}\)
A1 for \(73.4^\circ\)

Worked solution: \(\cos\theta = \dfrac{2}{|\mathbf{a}|} = \dfrac{2}{7}\), so \(\theta = 73.4^\circ\).

Question 2Medium5 marks

(a) Find the vector of magnitude 34 that is parallel to \(\mathbf{v} = -9\mathbf{i} + 12\mathbf{j} + 8\mathbf{k}\) and in the opposite direction to \(\mathbf{v}\).[2]

(b) The vector \(\mathbf{a} = 4\mathbf{i} + p\mathbf{j} - 5\mathbf{k}\), where \(p\) is a constant, has magnitude 21.
Find the possible values of \(p\).[3]

Show the answer and mark scheme

(a) Answer: \(18\mathbf{i} - 24\mathbf{j} - 16\mathbf{k}\)

M1 for \(|\mathbf{v}| = 17\) and \(-\frac{34}{17}\mathbf{v}\)
A1 for \(18\mathbf{i} - 24\mathbf{j} - 16\mathbf{k}\)

Worked solution: \(|\mathbf{v}| = \sqrt{(-9)^2 + 12^2 + 8^2} = 17\), so the vector is \(-\frac{34}{17}\mathbf{v} = -2(-9\mathbf{i} + 12\mathbf{j} + 8\mathbf{k}) = 18\mathbf{i} - 24\mathbf{j} - 16\mathbf{k}\)

(b) Answer: \(p = 20\) or \(p = -20\)

M1 for \(4^2 + (-5)^2 + p^2 = 21^2\)
A1 for \(p^2 = 400\)
A1 for \(p = 20\) and \(p = -20\)

Worked solution: \(4^2 + (-5)^2 + p^2 = 441 \Rightarrow p^2 = 400 \Rightarrow p = \pm 20\)

Question 3Hard9 marks

The vectors \(\mathbf{a}\) and \(\mathbf{b}\) are given by\[\mathbf{a} = -4\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}, \qquad \mathbf{b} = -\mathbf{i} - \mathbf{j} + \mathbf{k}\]and \(t\) is a real number.

(a) Show that \(|\mathbf{a} + t\mathbf{b}|^2 = 3t^{2} + 6t + 41\).[3]

(b) Find the values of \(t\) for which \(|\mathbf{a} + t\mathbf{b}| = \sqrt{41}\).[3]

(c) Find the value of \(t\) for which \(|\mathbf{a} + t\mathbf{b}|\) is least, and find this least value, giving it to 3 significant figures.[3]

Show the answer and mark scheme

(a) Answer: \(\mathbf{a} + t\mathbf{b} = (-4 - t)\mathbf{i} + (-3 - t)\mathbf{j} + (-4 + t)\mathbf{k}\), so \(|\mathbf{a} + t\mathbf{b}|^2 = 3t^{2} + 6t + 41\)

M1 for \(\mathbf{a} + t\mathbf{b} = (-4 - t)\mathbf{i} + (-3 - t)\mathbf{j} + (-4 + t)\mathbf{k}\)
M1 for \(|\mathbf{a} + t\mathbf{b}|^2 = (-4 - t)^2 + (-3 - t)^2 + (-4 + t)^2\)
A1* for expanding and simplifying to the given result with no errors

Worked solution: \(\mathbf{a} + t\mathbf{b} = (-4 - t)\mathbf{i} + (-3 - t)\mathbf{j} + (-4 + t)\mathbf{k}\)
\(|\mathbf{a} + t\mathbf{b}|^2 = (-4 - t)^2 + (-3 - t)^2 + (-4 + t)^2\)
\(= (t^{2} + 8t + 16) + (t^{2} + 6t + 9) + (t^{2} - 8t + 16) = 3t^{2} + 6t + 41\)

(b) Answer: \(t = -2\) or \(t = 0\)

M1 for \(3t^{2} + 6t + 41 = 41\)
dM1 for solving their 3-term quadratic
A1 for \(t = -2\) and \(t = 0\)

Worked solution: \(3t^{2} + 6t + 41 = 41 \Rightarrow 3t^{2} + 6t = 0 \Rightarrow t^{2} + 2t = 0\)
\(t(t + 2) = 0\), so \(t = -2\) or \(t = 0\)

M1 for completing the square or differentiating to find the minimum of \(3t^{2} + 6t + 41\)
A1 for \(t = -1\)
A1 for least value awrt 6.16 (exact value \(\sqrt{38}\))

Worked solution: \(3t^{2} + 6t + 41 = 3\left(t + 1\right)^2 + 38\)
The least value of \(|\mathbf{a} + t\mathbf{b}|^2\) is \(38\), when \(t = -1\), so the least value of \(|\mathbf{a} + t\mathbf{b}|\) is \(\sqrt{38} = \sqrt{38} = 6.16\) (3 s.f.)

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