The vector equation of line \(l\) is given as \(\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1 \\
  3 \\
  6
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}
  { - 1} \\
  2 \\
  { - 1}
\end{array}} \right)\) .

Find the Cartesian equation of the plane containing the line \(l\) and the point A(4, − 2, 5) .

EITHER

\(l\) goes through the point (1, 3, 6) , and the plane contains A(4, –2, 5)

the vector containing these two points is on the plane, i.e.

\(\left( {\begin{array}{*{20}{c}}
  1 \\
  3 \\
  6
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  4 \\
  { - 2} \\
  5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 3} \\
  5 \\
  1
\end{array}} \right)\)     (M1)A1

\(\left( {\begin{array}{*{20}{c}}
  { - 1} \\
  2 \\
  { - 1}
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  { - 3} \\
  5 \\
  1
\end{array}} \right) = \left| {\begin{array}{*{20}{c}}
  {\boldsymbol{i}}&{\boldsymbol{j}}&{\boldsymbol{k}} \\
  { - 1}&2&{ - 1} \\
  { - 3}&5&1
\end{array}} \right| = 7{\boldsymbol{i}} + 4{\boldsymbol{j}} + {\boldsymbol{k}}\)     M1A1

\(\left( {\begin{array}{*{20}{c}}
  4 \\
  { - 2} \\
  5
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  7 \\
  4 \\
  1
\end{array}} \right) = 25\)     (M1)

hence, Cartesian equation of the plane is \(7x + 4y + z = 25\)     A1

OR

finding a third point     M1

e.g. (0, 5, 5)     A1

three points are (1, 3, 6), (4, –2, 5), (0, 5, 5)

equation is \(ax + by + cz = 1\)

system of equations     M1

\(a + 3b + 6c = 1\)
\(4a - 2b + 5c = 1\)
\(5b + 5c = 1\)

\(a = \frac{7}{{25}}\) , \(b = \frac{4}{{25}}\) , \(c = \frac{1}{{25}}\) , from GDC     M1A1

so \(\frac{7}{{25}}x + \frac{4}{{25}}y + \frac{1}{{25}}z = 1\)     A1

or \(7x + 4y + z = 25\)

[6 marks]

There were many successful answers to this question, as would be expected. There seemed to be some students, however, that had not been taught the vector geometry section