| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Standard +0.3 This is a straightforward two-part vectors question requiring standard techniques: (i) using the dot product condition for perpendicularity (BA·BC = 0), and (ii) recognizing collinearity means vectors AB and BC are parallel. Both parts involve routine vector arithmetic with no conceptual challenges beyond applying well-practiced methods. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors |
2 Points $A , B$ and $C$ have position vectors $- 5 \mathbf { i } - 10 \mathbf { j } + 12 \mathbf { k } , \mathbf { i } + 2 \mathbf { j } - 3 \mathbf { k }$ and $3 \mathbf { i } + 6 \mathbf { j } + p \mathbf { k }$ respectively, where $p$ is a constant.\\
(i) Given that angle $A B C = 90 ^ { \circ }$, find the value of $p$.\\
(ii) Given instead that $A B C$ is a straight line, find the value of $p$.
\hfill \mbox{\textit{OCR C4 2010 Q2 [6]}}