Date | May 2013 | Marks available | 4 | Reference code | 13M.2.hl.TZ1.11 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Show that | Question number | 11 | Adapted from | N/A |
Question
Consider the points P(−3, −1, 2) and Q(5, 5, 6).
Find a vector equation for the line, L1L1, which passes through the points P and Q.
The line L2L2 has equation
r=(−404)+s(520).r=⎛⎜⎝−404⎞⎟⎠+s⎛⎜⎝520⎞⎟⎠.
Show that L1L1 and L2L2 intersect at the point R(1, 2, 4).
Find the acute angle between L1L1 and L2L2.
Let S be a point on L2L2 such that |→RP|=|→RS|∣∣∣−−→RP∣∣∣=∣∣∣−→RS∣∣∣.
Show that one of the possible positions for S is S1S1(−4, 0, 4) and find the coordinates of the other possible position, S2S2.
Let S be a point on L2L2 such that |→RP|=|→RS|∣∣∣−−→RP∣∣∣=∣∣∣−→RS∣∣∣.
Find a vector equation of the line which passes through R and bisects PˆRS1P^RS1.
Markscheme
→PQ=(864)−−→PQ=⎛⎜⎝864⎞⎟⎠ (A1)
equation of line: r=(−3−12)+t(864)r=⎛⎜⎝−3−12⎞⎟⎠+t⎛⎜⎝864⎞⎟⎠ (or equivalent) M1A1
Note: Award M1A0 if r = is omitted.
[3 marks]
METHOD 1
x: −4+5s=−3+8tx: −4+5s=−3+8t
y: 2s=−1+6ty: 2s=−1+6t
z: 4=2+4tz: 4=2+4t M1
solving any two simultaneously M1
t = 0.5, s = 1 (or equivalent) A1
verification that these values give R when substituted into both equations (or that the three equations are consistent and that one gives R) R1
METHOD 2
(1, 2, 4) is given by t = 0.5 for L1L1 and s = 1 for L2L2 M1A1A1
because (1, 2, 4) is on both lines it is the point of intersection of the two lines R1
[4 marks]
(520)⋅(432)=26=√29×√29cosθ⎛⎜⎝520⎞⎟⎠⋅⎛⎜⎝432⎞⎟⎠=26=√29×√29cosθ M1
cosθ=2629cosθ=2629 (A1)
θ=0.459θ=0.459 or 26.3° A1
[3 marks]
→RP=(−3−12)−(124)=(−4−3−2)−−→RP=⎛⎜⎝−3−12⎞⎟⎠−⎛⎜⎝124⎞⎟⎠=⎛⎜⎝−4−3−2⎞⎟⎠, |→RP|=√29∣∣∣−−→RP∣∣∣=√29 (M1)A1
Note: This could also be obtained from |0.5(864)|∣∣ ∣∣0.5⎛⎜⎝864⎞⎟⎠∣∣ ∣∣
EITHER
→RS1=(−404)−(124)=(−5−20)−−→RS1=⎛⎜⎝−404⎞⎟⎠−⎛⎜⎝124⎞⎟⎠=⎛⎜⎝−5−20⎞⎟⎠, |→RS1|=√29∣∣∣−−→RS1∣∣∣=√29 A1
∴→OS2=→OS1+2→S1R=(−404)+2(520) M1A1
(or →OS2=→OR+→S1R=(124)+(520))
=(644)
S2 is (6, 4, 4) A1
OR
(−4+5s2s4)−(124)=(5s−52s−20) M1
(5s−5)2+(2s−2)2=29 M1A1
29s2−58s+29=29
s(s−2)=0, s=0, 2
(6, 4, 4) (and (−4, 0, 4)) A1
Note: There are several geometrical arguments possible using information obtained in previous parts, depending on what forms the previous answers had been given.
[6 marks]
EITHER
midpoint of [PS1] is M(–3.5, –0.5, 3) M1A1
→RM=(−4.5−2.5−1) A1
OR
→RS1=(−5−20) M1
the direction of the line is →RS1+→RP
(−5−20)+(−4−3−2)=(−9−5−2) M1A1
THEN
the equation of the line is:
r=(124)+t(952) or equivalent A1
Note: Marks cannot be awarded for methods involving halving the angle, unless it is clear that the candidate considers also the equation of the plane of L1 and L2 to reduce the number of parameters involved to one (to obtain the vector equation of the required line).
[4 marks]
Examiners report
There were many good answers to part (a) showing a clear understanding of finding the vector equation of a line. Unfortunately this understanding was marred by many students failing to write the equation properly resulting in just 2 marks out of the 3. The most common response was of the form L1=(−3−12)+λ(432) which seemed a waste of a mark.
In part (b) many students failed to verify that the lines do indeed intersect.
Part (c) was very well done.
In part (d) most candidates were able to obtain the first three marks, but few were able to find the second point.
There were few correct answers to part (e).