Edexcel A level Maths (9MA0) · Pure mathematics › Vectors
Practise Vectors in 2D and 3D. unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
The vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) are given by\[\mathbf{a} = \mathbf{i} + p\mathbf{j} - \mathbf{k}, \qquad \mathbf{b} = q\mathbf{i} + \mathbf{j} - 6\mathbf{k}, \qquad \mathbf{c} = 9\mathbf{i} - 9\mathbf{j} + r\mathbf{k}\]where \(p\), \(q\) and \(r\) are constants.
(a) Given that \(3\mathbf{a} - 3\mathbf{b} = \mathbf{c}\), find the values of \(p\), \(q\) and \(r\).[3]
(b) Using your values of \(p\) and \(q\), find the values of the constants \(\lambda\) and \(\mu\) such that\[\lambda\mathbf{a} + \mu\mathbf{b} = -3\mathbf{i} - 3\mathbf{j} - 21\mathbf{k}\][4]
Show the answer and mark scheme
(a)Answer: \(p = -2, \ q = -2, \ r = 15\)
M1 for equating the coefficients of \(\mathbf{i}\), \(\mathbf{j}\) or \(\mathbf{k}\) to form a correct equation, e.g. \(3p - 3 = -9\)
A1 for two of \(p = -2\), \(q = -2\), \(r = 15\)
A1 for all three correct
Worked solution: \(\mathbf{i}\): \(3 - 3q = 9\) so \(q = -2\) \(\mathbf{j}\): \(3p - 3 = -9\) so \(p = -2\) \(\mathbf{k}\): \(3 \times (-1) - 3 \times (-6) = r\) so \(r = 15\)
(b)Answer: \(\lambda = 3, \ \mu = 3\)
M1 for equating components to form at least two equations in \(\lambda\) and \(\mu\), e.g. \(\lambda - 2\mu = -3\)
dM1 for solving two of the equations simultaneously
A1 for \(\lambda = 3\)
A1 for \(\mu = 3\)
Worked solution: Equating the \(\mathbf{i}\) and \(\mathbf{j}\) components: \(\lambda - 2\mu = -3 \quad (1)\) \(-2\lambda + \mu = -3 \quad (2)\) \((1) + 2 \times (2)\): \(-3\lambda = -9\), so \(\lambda = 3\). Substituting into (1): \(\mu = 3\). Check with \(\mathbf{k}\): \(3 \times (-1) + 3 \times (-6) = -21\) ✓
Question 3Hard8 marks
The vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) are given by\[\mathbf{a} = 4\mathbf{i} + 5\mathbf{j} + 4\mathbf{k}, \qquad \mathbf{b} = 2\mathbf{i} + 4\mathbf{j} - \mathbf{k}, \qquad \mathbf{c} = -12\mathbf{i} - 18\mathbf{j} - 4\mathbf{k}\]
(a) Show that there are no real numbers \(\lambda\) and \(\mu\) such that \(\lambda\mathbf{a} + \mu\mathbf{b} = \mathbf{c}\).[4]
(b) The vector \(\mathbf{d} = 6\mathbf{i} + p\mathbf{j} + 3\mathbf{k}\), where \(p\) is a constant, can be written in the form \(\lambda\mathbf{a} + \mu\mathbf{b}\). Find the value of \(p\).[4]
Show the answer and mark scheme
(a)Answer: The \(\mathbf{i}\) and \(\mathbf{j}\) equations give \(\lambda = -2, \ \mu = -2\), but then the \(\mathbf{k}\) component is \(-6 \ne -4\), so no such \(\lambda\) and \(\mu\) exist.
M1 for equating components to form equations in \(\lambda\) and \(\mu\), e.g. \(4\lambda + 2\mu = -12\) and \(5\lambda + 4\mu = -18\)
dM1 for solving two of the equations simultaneously
A1 for \(\lambda = -2\) and \(\mu = -2\) (from the \(\mathbf{i}\) and \(\mathbf{j}\) equations) or correct values from another pair
A1* for substituting into the remaining equation, showing it is not satisfied, e.g. \(-2 \times 4 - 2 \times (-1) = -6 \ne -4\), and concluding that no such \(\lambda\) and \(\mu\) exist
Worked solution: Equating the \(\mathbf{i}\) and \(\mathbf{j}\) components: \(4\lambda + 2\mu = -12 \quad (1)\) \(5\lambda + 4\mu = -18 \quad (2)\) \(2 \times (1) - (2)\): \(3\lambda = -6\), so \(\lambda = -2\). Substituting into (1): \(\mu = -2\). \(\mathbf{k}\) component: \(-2 \times 4 - 2 \times (-1) = -6\), but the \(\mathbf{k}\) component of \(\mathbf{c}\) is \(-4\). The three equations are inconsistent, so there are no real numbers \(\lambda\) and \(\mu\) with \(\lambda\mathbf{a} + \mu\mathbf{b} = \mathbf{c}\).