| Date | November 2016 | Marks available | 4 | Reference code | 16N.1.hl.TZ0.8 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 8 | Adapted from | N/A |
Question
Consider the lines l1l1 and l2l2 defined by
l1:l1: r =(−3−2a)+β(142)=⎛⎜⎝−3−2a⎞⎟⎠+β⎛⎜⎝142⎞⎟⎠ and l2:6−x3=y−24=1−zl2:6−x3=y−24=1−z where aa is a constant.
Given that the lines l1l1 and l2l2 intersect at a point P,
find the value of aa;
determine the coordinates of the point of intersection P.
Markscheme
METHOD 1
l1:l1:r =(−3−2a)=β(142)⇒{x=−3+βy=−2+4βz=a+2β=⎛⎜⎝−3−2a⎞⎟⎠=β⎛⎜⎝142⎞⎟⎠⇒⎧⎪⎨⎪⎩x=−3+βy=−2+4βz=a+2β M1
6−(−3+β)3=(−2+4β)−24⇒4=4β3⇒β=36−(−3+β)3=(−2+4β)−24⇒4=4β3⇒β=3 M1A1
6−(−3+β)3=1−(a+2β)⇒2=−5−a⇒a=−76−(−3+β)3=1−(a+2β)⇒2=−5−a⇒a=−7 A1
METHOD 2
{−3+β=6−3λ−2+4β=4λ+2a+2β=1−λ⎧⎪⎨⎪⎩−3+β=6−3λ−2+4β=4λ+2a+2β=1−λ M1
attempt to solve M1
λ=2, β=3λ=2, β=3 A1
a=1−λ−2β=−7a=1−λ−2β=−7 A1
[4 marks]
→OP=(−3−2−7)+3∙(142)−−→OP=⎛⎜⎝−3−2−7⎞⎟⎠+3∙⎛⎜⎝142⎞⎟⎠ (M1)
=(010−1)=⎛⎜⎝010−1⎞⎟⎠ A1
∴P(0, 10, −1)
[2 marks]
