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1.10 Vectors

1.10 Vectors

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Question 29

Relative to a fixed origin OOO, the point A A\,A has position vector (4i−j+3k)(4\mathbf{i} - \mathbf{j} + 3\mathbf{k})(4i−j+3k), the point B B\,B has position vector (2i+3j−k)(2\mathbf{i} + 3\mathbf{j} - \mathbf{k})(2i+3j−k), and the point C C\,C has position vector (ai+2j+5k)(a\mathbf{i} + 2\mathbf{j} + 5\mathbf{k})(ai+2j+5k), 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.

a.

Find the position vector of DDD.

[2]
b.

Given that ∣AC⃗∣=7|\vec{AC}| = 7∣AC∣=7, find the value of aaa.

[3]
Markscheme

1.10 Vectors Questions

  1. A Level
  2. /Maths
  3. /1.10 Vectors

197 exam-style questions on OCR A Level Maths 1.10 Vectors, covering 1.10.1 Vectors in two dimensions, 1.10.2 Vectors in three dimensions (A-level only), 1.10.3 Magnitude and direction of vectors, 1.10.4 Basic operations on vectors, 1.10.5 Position vectors, 1.10.6 Distance between points, 1.10.7 Problem solving using vectors, 1.10.8 Vectors in kinematics, and 1.10 Vectors. Each one has a worked solution and a mark scheme showing where the marks go.

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