Consider the vectors a = 6i + 3j + 2k, b = −3j + 4k.

(i)     Find the cosine of the angle between vectors a and b.

(ii)     Find a \( \times \) b.

(iii)     Hence find the Cartesian equation of the plane \(\prod \) containing the vectors a and b and passing through the point (1, 1, −1).

(iv)     The plane \(\prod \) intersects the x-y plane in the line l. Find the area of the finite triangular region enclosed by l, the x-axis and the y-axis.

Given two vectors p and q,

(i)     show that p\( \cdot \)p = \(|\)p\({|^2}\);

(ii)     hence, or otherwise, show that \(|\)p + q\({|^2}\) = \(|\)p\({|^2}\) + 2p\( \cdot \)q + \(|\)q\({|^2}\);

(iii)     deduce that \(|\)p + q\(|\) ≤ \(|\)p\(|\) + \(|\)q\(|\).

(i)     use of a\( \cdot \)b = \(|\)a\(|\)\(|\)b\(|\cos \theta \)     (M1)

a\( \cdot \)b = –1     (A1)

\(|\)a\(|\) = 7, \(|\)b\(|\) = 5     (A1)

\(\cos \theta = - \frac{1}{{35}}\)     A1

(ii)     the required cross product is

\(\left| {\begin{array}{*{20}{c}}
  i&j&k \\
  6&3&2 \\
  0&{ - 3}&4
\end{array}} \right| = \) 18i - 24j -18k     M1A1

(iii)     using r\( \cdot \)n = p\( \cdot \)n the equation of the plane is     (M1)

\(18x - 24y - 18z = 12\,\,\,\,\,(3x - 4y - 3z = 2)\)     A1

(iv)     recognizing that z = 0     (M1)

x-intercept \( = \frac{2}{3}\), y-intercept \( = - \frac{1}{2}\)     (A1)

area \( = \left( {\frac{2}{3}} \right)\left( {\frac{1}{2}} \right)\left( {\frac{1}{2}} \right) = \frac{1}{6}\)     A1

[11 marks]

(i)     p\( \cdot \)p = \(|\)p\(|\)\(|\)p\(|\cos 0\)     M1A1

= \(|\)p\({|^2}\)     AG

(ii)     consider the LHS, and use of result from part (i)

\(|\)p + q\({|^2}\) = (p + q)\( \cdot \)(p + q)     M1

= p\( \cdot \)p + p\( \cdot \)q + q\( \cdot \)p + q\( \cdot \)q     (A1)

= p\( \cdot \)p + 2p\( \cdot \)q + q\( \cdot \)q     A1

= \(|\)p\({|^2}\) + 2p\( \cdot \)q + \(|\)q\({|^2}\)     AG

(iii)     EITHER

use of p\( \cdot \)q \( \leqslant \) \(|\)p\(|\)\(|\)q\(|\)     M1

so 0 \( \leqslant \) \(|\)p + q\({|^2}\) = \(|\)p\({|^2}\) + 2p\( \cdot \)q + \(|\)q\({|^2}\) \( \leqslant \) \(|\)p\({|^2}\) + 2 \(|\)p\(|\)\(|\)q\(|\) + \(|\)q\({|^2}\)     A1

take square root (of these positive quantities) to establish     A1

\(|\)p + q\(|\) \( \leqslant \) \(|\)p\(|\) + \(|\)q\(|\)     AG

OR

    M1M1

Note: Award M1 for correct diagram and M1 for correct labelling of vectors including arrows.

since the sum of any two sides of a triangle is greater than the third side,

\(|\)p\(|\) + \(|\)q\(|\) > \(|\)p + q\(|\)     A1

when p and q are collinear \(|\)p\(|\) + \(|\)q\(|\) = \(|\)p + q\(|\)

\( \Rightarrow |\)p + q\(|\) \( \leqslant \) \(|\)p\(|\) + \(|\)q\(|\)     AG

[8 marks]

This was the most accessible question in section B for the candidates. The majority of candidates produced partially correct answers to part (a), with nearly all candidates being able to use the scalar and vector product. Candidates found part (iv) harder and often did not appreciate the significance of letting z = 0. Candidates clearly found part (b) harder and again this was a point where candidates lost time. Many candidates attempted this using components, which was fine in part (i), fine, but time consuming in part (ii), and extremely complicated in part (iii). A number of candidates lost marks because they were careless in showing their working in part (ii) which required them to “show that”.

This was the most accessible question in section B for the candidates. The majority of candidates produced partially correct answers to part (a), with nearly all candidates being able to use the scalar and vector product. Candidates found part (iv) harder and often did not appreciate the significance of letting z = 0. Candidates clearly found part (b) harder and again this was a point where candidates lost time. Many candidates attempted this using components, which was fine in part (i), fine, but time consuming in part (ii), and extremely complicated in part (iii). A number of candidates lost marks because they were careless in showing their working in part (ii) which required them to “show that”.