Date | November 2013 | Marks available | 5 | Reference code | 13N.2.sl.TZ0.9 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Show that | Question number | 9 | Adapted from | N/A |
Question
Consider the lines L1 and L2 with equations L1 : r=(1182)+s(43−1) and L2 : r=(11−7)+t(2111).
The lines intersect at point P.
Find the coordinates of P.
Show that the lines are perpendicular.
The point Q(7,5,3) lies on L1. The point R is the reflection of Q in the line L2.
Find the coordinates of R.
Markscheme
appropriate approach (M1)
eg (1182)+s(43−1)=(11−7)+t(2111), L1=L2
any two correct equations A1A1
eg 11+4s=1+2t, 8+3s=1+t, 2−s=−7+11t
attempt to solve system of equations (M1)
eg 10+4s=2(7+3s),{4s−2t=−103s−t=−7
one correct parameter A1
eg s=−2, t=1
P(3,2,4) (accept position vector) A1 N3
[6 marks]
choosing correct direction vectors for L1 and L2 (A1)(A1)
eg (43−1),(2111) (or any scalar multiple)
evidence of scalar product (with any vectors) (M1)
eg a⋅b, (43−1)∙(2111)
correct substitution A1
eg 4(2)+3(1)+(−1)(11), 8+3−11
calculating a⋅b=0 A1
Note: Do not award the final A1 without evidence of calculation.
vectors are perpendicular AG N0
[5 marks]
Note: Candidates may take different approaches, which do not necessarily involve vectors.
In particular, most of the working could be done on a diagram. Award marks in line with the markscheme.
METHOD 1
attempt to find →QP or →PQ (M1)
correct working (may be seen on diagram) A1
eg →QP = (−4−31), →PQ = (753)−(324)
recognizing R is on L1 (seen anywhere) (R1)
eg on diagram
Q and R are equidistant from P (seen anywhere) (R1)
eg →QP=→PR, marked on diagram
correct working (A1)
eg (324)−(753)=(xyz)−(324),(−4−31)+(324)
{\text{R}}(–1, –1, 5) (accept position vector) A1 N3
METHOD 2
recognizing {\text{R}} is on {L_1} (seen anywhere) (R1)
eg on diagram
{\text{Q}} and {\text{R}} are equidistant from {\text{P}} (seen anywhere) (R1)
eg {\text{P}} midpoint of {\text{QR}}, marked on diagram
valid approach to find one coordinate of mid-point (M1)
eg {x_p} = \frac{{{x_Q} + {x_R}}}{2},{\text{ }}2{y_p} = {y_Q} + {y_R},{\text{ }}\frac{1}{2}\left( {{z_Q} + {z_R}} \right)
one correct substitution A1
eg {x_R} = 3 + (3 - 7),{\text{ }}2 = \frac{{5 + {y_R}}}{2},{\text{ }}4 = \frac{1}{2}(z + 3)
correct working for one coordinate (A1)
eg {x_R} = 3 - 4,{\text{ }}4 - 5 = {y_R},{\text{ }}8 = (z + 3)
{\text{R}} (-1, -1, 5) (accept position vector) A1 N3
[6 marks]