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