Relative to a fixed origin OOO, the point A A\,A has position vector (2i−j+5k)(2\mathbf{i} - \mathbf{j} + 5\mathbf{k})(2i−j+5k), the point B B\,B has position vector (5i+2j+k)(5\mathbf{i} + 2\mathbf{j} + \mathbf{k})(5i+2j+k), and the point C C\,C has position vector (i+aj+3k)(\mathbf{i} + a\mathbf{j} + 3\mathbf{k})(i+aj+3k), where a a\,a is a constant and a>0a > 0a>0. D D\,D is the point such that AB→=BD→\overrightarrow{AB} = \overrightarrow{BD}AB=BD.
Find the position vector of DDD.
Given that ∣AC⃗∣=21|\vec{AC}| = \sqrt{21}∣AC∣=21, find the value of aaa.
168 exam-style questions on OCR (MEI) A Level Maths 1.11 Vectors, covering 1.11.1 Language of vectors in two dimensions, 1.11.2 Add, subtract and scale vectors, 1.11.3 Magnitude and direction of a vector, 1.11.4 Position vectors, 1.11.5 Distance between points by position vectors, 1.11.6 Vectors to solve problems, 1.11.7 Language of vectors in three dimensions (A-level only), and 1.11 Vectors. Each one has a worked solution and a mark scheme showing where the marks go.