Three forces act on a particle:
F1=(5i−2j+3k)\mathbf{F}_1 = (5\mathbf{i} - 2\mathbf{j} + 3\mathbf{k})F1=(5i−2j+3k) N, F2=(−3i+7j−k)\mathbf{F}_2 = (-3\mathbf{i} + 7\mathbf{j} - \mathbf{k})F2=(−3i+7j−k) N, F3=(ai+bj−2k)\mathbf{F}_3 = (a\mathbf{i} + b\mathbf{j} - 2\mathbf{k})F3=(ai+bj−2k) N
where a a\,a and b b\,b are constants.
Given that the particle is in equilibrium, find the value of a a\,a and the value of bbb.
The force F3\mathbf{F}_3F3 is now removed. Find the magnitude of the resultant of F1\mathbf{F}_1F1 and F2\mathbf{F}_2F2, giving your answer in newtons to three significant figures.
168 exam-style questions on AQA A Level Maths 1.13 J: Vectors, covering 1.13.1 Vectors in two and three dimensions, 1.13.2 Magnitude and direction of a vector, 1.13.3 Vector arithmetic, 1.13.4 Position vectors, and 1.13.5 Vectors to solve problems. Each one has a worked solution and a mark scheme showing where the marks go.