Consider the points A(1, −1, 4), B (2, − 2, 5) and O(0, 0, 0).

(a)     Calculate the cosine of the angle between $\overrightarrow {{\text{OA}}}$ and $\overrightarrow {{\text{AB}}}$.

(b)     Find a vector equation of the line ${L_1}$ which passes through A and B.

The line ${L_2}$ has equation r = 2i + 4j + 7k + t(2i + j + 3k), where $t \in \mathbb{R}$ .

(c)     Show that the lines ${L_1}$ and ${L_2}$ intersect and find the coordinates of their point of intersection.

(d)     Find the Cartesian equation of the plane which contains both the line ${L_2}$ and the point A.

(a)     Use of $\cos \theta  = \frac{{\overrightarrow {{\text{OA}}}  \cdot \overrightarrow {{\text{AB}}} }}{{\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{AB}}} } \right|}}$     (M1)

${\overrightarrow {{\text{AB}}} }$ = i − j + k     A1

$\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt 3$ and $\left| {\overrightarrow {{\text{OA}}} } \right| = 3\sqrt 2$     A1

$\overrightarrow {{\text{OA}}}  \cdot \overrightarrow {{\text{AB}}} = 6$     A1

substituting gives $\cos \theta = \frac{2}{{\sqrt 6 }}\,\,\,\,\,\left( { = \frac{{\sqrt 6 }}{3}} \right)\,\,\,\,\,$or equivalent     M1     N1

[5 marks]

 

(b)     ${L_1}$ : r = $\overrightarrow {{\text{OA}}} + s\overrightarrow {{\text{AB}}} \,\,\,\,\,$ or equivalent     (M1)

${L_1}$ : r = i − j + 4k + s(i − j + k)$\,\,\,\,\,$or equivalent     A1

Note: Award (M1)A0 for omitting “ r = ” in the final answer.

 

[2 marks]

 

(c)     Equating components and forming equations involving s and t     (M1)

1 + s = 2 + 2t , −1 − s = 4 + t , 4 + s = 7 + 3t

Having two of the above three equations     A1A1

Attempting to solve for s or t     (M1)

Finding either s = −3 or t = −2     A1

Explicitly showing that these values satisfy the third equation     R1

Point of intersection is (−2, 2, 1)     A1     N1

Note: Position vector is not acceptable for final A1.

 

[7 marks]

 

(d)     METHOD 1

$r = \left( {\begin{array}{*{20}{c}}   1 \\   { - 1} \\   4 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}   2 \\   1 \\   3 \end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}   { - 3} \\   3 \\   { - 3} \end{array}} \right)$     (A1)

$x = 1 + 2\lambda - 3\mu {\text{ , }}y = - 1 + \lambda + 3\mu {\text{ and }}z = 4 + 3\lambda - 3\mu$     M1A1

Elimination of the parameters     M1

$x + y = 3\lambda {\text{ so }}4(x + y) = 12\lambda {\text{ and }}y + z = 4\lambda + 3{\text{ so }}3(y + z) = 12\lambda + 9$

$3(y + z) = 4(x + y) + 9$     A1

Cartesian equation of plane is 4x + y − 3z = −9 (or equivalent)     A1     N1

[6 marks] 

METHOD 2

EITHER

The point (2, 4, 7) lies on the plane.

The vector joining (2, 4, 7) and (1, −1, 4) and 2i + j + 3k are parallel to the plane. So they are perpendicular to the normal to the plane.

(i − j + 4k) − (2i + 4j + 7k) = −i − 5j − 3k     (A1)

$n = \left| {\begin{array}{*{20}{c}}   i&j&k \\   { - 1}&{ - 5}&{ - 3} \\   2&1&3 \end{array}} \right|$     M1

= −12i − 3j + 9k$\,\,\,\,\,$or equivalent parallel vector     A1

OR

${L_1}$ and ${L_2}$ intersect at D(−2, 2, 1)

$n = \left| {\begin{array}{*{20}{c}}   i&j&k \\   2&1&{ - 3} \\   { - 3}&3&{ - 3} \end{array}} \right|$     M1

= −12i − 3j + 9k$\,\,\,\,\,$or equivalent parallel vector     A1

THEN

r$\cdot$n = (i − j + 4k)$\cdot$(−12i − 3j + 9k)     M1

= 27     A1

Cartesian equation of plane is 4x + y −3z = −9 (or equivalent)     A1     N1

[6 marks]

Total [20 marks]

Most candidates scored reasonably well on this question. The most common errors were: Using OB rather than AB in (a); omitting the r = in (b); failure to check that the values of the two parameters satisfied the third equation in (c); the use of an incorrect vector in (d). Even when (d) was correctly answered, there was usually little evidence of why a specific vector had been used.