Find a vector equation for L₁.

The vector $\left( \begin{gathered} 2 \hfill \\ p \hfill \\ 0 \hfill \\ \end{gathered} \right)$ is perpendicular to $\mathop {{\text{AB}}}\limits^ \to$. Find the value of p.

any correct equation in the form r = a + tb (accept any parameter for t)

where a is $\left( \begin{gathered} 2 \hfill \\ 1 \hfill \\ 3 \hfill \\ \end{gathered} \right)$, and b is a scalar multiple of $\left( \begin{gathered} 1 \hfill \\ 3 \hfill \\ 1 \hfill \\ \end{gathered} \right)$     A2 N2

eg r = $\left( \begin{gathered} 2 \hfill \\ 1 \hfill \\ 3 \hfill \\ \end{gathered} \right) = t\left( \begin{gathered} 1 \hfill \\ 3 \hfill \\ 1 \hfill \\ \end{gathered} \right)$, r = 2i + j + 3k + s(i + 3j + k)

Note: Award A1 for the form a + tb, A1 for the form L = a + tb, A0 for the form r = b + ta.

[2 marks]

METHOD 1

correct scalar product     (A1)

eg  (1 × 2) + (3 × p) + (1 × 0), 2 + 3p

evidence of equating their scalar product to zero     (M1)

eg  a•b = 0, 2 + 3p = 0, 3p = −2

$p =  - \frac{2}{3}$       A1 N3

METHOD 2

valid attempt to find angle between vectors      (M1)

correct substitution into numerator and/or angle       (A1)

eg  ${\text{cos}}\,\theta  = \frac{{\left( {1 \times 2} \right) + \left( {3 \times p} \right) + \left( {1 \times 0} \right)}}{{\left| a \right|\left| b \right|}},\,\,{\text{cos}}\,\theta  = 0$

$p =  - \frac{2}{3}$       A1 N3

[3 marks]