SPS SPS SM Pure 2023 June — Question 4 6 marks

Exam BoardSPS
ModuleSPS SM Pure (SPS SM Pure)
Year2023
SessionJune
Marks6
TopicVectors Introduction & 2D
TypeRatio division of line segment
DifficultyModerate -0.8 This is a straightforward vectors question testing basic operations: finding magnitude using Pythagoras (part a), using parallelogram properties with vector addition (part b), and section formula/ratio theorem for collinear points (part c). All three parts are standard textbook exercises requiring only routine application of well-practiced techniques with no problem-solving insight needed.
Spec1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors

Relative to a fixed origin \(O\), • the point \(A\) has position vector \(5\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}\) • the point \(B\) has position vector \(7\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) • the point \(C\) has position vector \(4\mathbf{i} + 8\mathbf{j} - 3\mathbf{k}\)
  1. Find \(|\vec{AB}|\) giving your answer as a simplified surd. [2] Given that \(ABCD\) is a parallelogram,
  2. find the position vector of the point \(D\). [2] The point \(E\) is positioned such that • \(ACE\) is a straight line • \(AC : CE = 2 : 1\)
  3. Find the coordinates of the point \(E\). [2]

Relative to a fixed origin $O$,

• the point $A$ has position vector $5\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}$

• the point $B$ has position vector $7\mathbf{i} + \mathbf{j} + 2\mathbf{k}$

• the point $C$ has position vector $4\mathbf{i} + 8\mathbf{j} - 3\mathbf{k}$

(a) Find $|\vec{AB}|$ giving your answer as a simplified surd. [2]

Given that $ABCD$ is a parallelogram,

(b) find the position vector of the point $D$. [2]

The point $E$ is positioned such that

• $ACE$ is a straight line

• $AC : CE = 2 : 1$

(c) Find the coordinates of the point $E$. [2]

\hfill \mbox{\textit{SPS SPS SM Pure 2023 Q4 [6]}}