IB Questionbank

User interface language: English | Español

Date	May 2018	Marks available	2	Reference code	18M.1.SL.TZ1.S_9
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Find	Question number	S_9	Adapted from	N/A

Question

Point A has coordinates (−4, −12, 1) and point B has coordinates (2, −4, −4).

The line L passes through A and B.

Show that $\mathop {{\text{AB}}}\limits^ \to = \left( \begin{gathered} \,6 \hfill \\ \,8 \hfill \\ - 5 \hfill \\ \end{gathered} \right)$

[1]

Find a vector equation for L.

[2]

b.i.

Point C (k , 12 , −k) is on L. Show that k = 14.

[4]

b.ii.

Find $\mathop {{\text{OB}}}\limits^ \to \, \bullet \mathop {{\text{AB}}}\limits^ \to$ .

[2]

c.i.

Write down the value of angle OBA.

[1]

c.ii.

Point D is also on L and has coordinates (8, 4, −9).

Find the area of triangle OCD.

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

correct approach A1

eg $\mathop {{\text{AO}}}\limits^ \to \,\, + \,\,\mathop {{\text{OB}}}\limits^ \to ,\,\,\,{\text{B}} - {\text{A}}\,{\text{, }}\,\left( \begin{gathered} \,\,2 \hfill \\ - 4 \hfill \\ - 4 \hfill \\ \end{gathered} \right) - \left( \begin{gathered} \, - 4 \hfill \\ - 12 \hfill \\ \,\,\,1 \hfill \\ \end{gathered} \right)$

$\mathop {{\text{AB}}}\limits^ \to = \left( \begin{gathered} \,6 \hfill \\ \,8 \hfill \\ - 5 \hfill \\ \end{gathered} \right)$ AG N0

[1 mark]

any correct equation in the form r = a + tb (any parameter for t) A2 N2

where a is $\left( \begin{gathered} \,\,2 \hfill \\ - 4 \hfill \\ - 4 \hfill \\ \end{gathered} \right)$ or $\left( \begin{gathered} \, - 4 \hfill \\ - 12 \hfill \\ \,\,\,1 \hfill \\ \end{gathered} \right)$ and b is a scalar multiple of $\left( \begin{gathered} \,6 \hfill \\ \,8 \hfill \\ - 5 \hfill \\ \end{gathered} \right)$

eg r $= \left( \begin{gathered} \, - 4 \hfill \\ - 12 \hfill \\ \,\,\,1 \hfill \\ \end{gathered} \right) + t\left( \begin{gathered} \,6 \hfill \\ \,8 \hfill \\ - 5 \hfill \\ \end{gathered} \right),\,\,\left( {x,\,\,y,\,\,z} \right) = \left( {2,\,\, - 4,\,\, - 4} \right) + t\left( {6,\,\,8,\,\, - 5} \right),$ r $= \left( \begin{gathered} \, - 4 + 6t \hfill \\ - 12 + 8t \hfill \\ \,\,\,1 - 5t \hfill \\ \end{gathered} \right)$

Note: Award A1 for the form a + tb, A1 for the form L = a + tb, A0 for the form r = b + ta.

[2 marks]

b.i.

METHOD 1 (solving for t)

valid approach (M1)

eg $\left( \begin{gathered} \,k \hfill \\ 12 \hfill \\ - k \hfill \\ \end{gathered} \right) = \left( \begin{gathered} \,\,2 \hfill \\ - 4 \hfill \\ - 4 \hfill \\ \end{gathered} \right) + t\left( \begin{gathered} \,6 \hfill \\ \,8 \hfill \\ - 5 \hfill \\ \end{gathered} \right),\,\,\left( \begin{gathered} \,k \hfill \\ 12 \hfill \\ - k \hfill \\ \end{gathered} \right) = \left( \begin{gathered} \, - 4 \hfill \\ - 12 \hfill \\ \,\,\,1 \hfill \\ \end{gathered} \right) + t\left( \begin{gathered} \,6 \hfill \\ \,8 \hfill \\ - 5 \hfill \\ \end{gathered} \right)$

one correct equation A1

eg −4 + 8t = 12, −12 + 8t = 12

correct value for t (A1)

eg t = 2 or 3

correct substitution A1

eg 2 + 6(2), −4 + 6(3), −[1 + 3(−5)]

k = 14 AG N0

METHOD 2 (solving simultaneously)

valid approach (M1)

two correct equations in A1

eg k = −4 + 6t, −k = 1 −5t

EITHER (eliminating k)

correct value for t (A1)

eg t = 2 or 3

correct substitution A1

eg 2 + 6(2), −4 + 6(3)

OR (eliminating t)

correct equation(s) (A1)

eg 5k + 20 = 30t and −6k − 6 = 30t, −k = 1 − 5 $\left( {\frac{{k + 4}}{6}} \right)$

correct working clearly leading to k = 14 A1

eg −k + 14 = 0, −6k = 6 −5k − 20, 5k = −20 + 6(1 + k)

THEN

k = 14 AG N0

[4 marks]

b.ii.

correct substitution into scalar product A1

eg (2)(6) − (4)(8) − (4)(−5), 12 − 32 + 20

$\mathop {{\text{OB}}}\limits^ \to \, \bullet \mathop {{\text{AB}}}\limits^ \to$ = 0 A1 N0

[2 marks]

c.i.

${\text{O}}\mathop {\text{B}}\limits^ \wedge {\text{A}} = \frac{\pi }{2},\,\,90^\circ \,\,\,\,\,\left( {{\text{accept}}\,\frac{{3\pi }}{2},\,\,270^\circ } \right)\,$ A1 N1

[1 marks]

c.ii.

METHOD 1 ( $\frac{1}{2}$ × height × CD)

recognizing that OB is altitude of triangle with base CD (seen anywhere) M1

eg $\frac{1}{2} \times \left| {\mathop {{\text{OB}}}\limits^ \to } \right| \times \left| {\mathop {{\text{CD}}}\limits^ \to } \right|,\,\,{\text{OB}} \bot {\text{CD}},$ sketch showing right angle at B

$\mathop {{\text{CD}}}\limits^ \to = \left( \begin{gathered} - 6 \hfill \\ - 8 \hfill \\ \,5 \hfill \\ \end{gathered} \right)$ or $\mathop {{\text{DC}}}\limits^ \to = \left( \begin{gathered} \,6 \hfill \\ \,8 \hfill \\ - 5 \hfill \\ \end{gathered} \right)$ (seen anywhere) (A1)

correct magnitudes (seen anywhere) (A1)(A1)

$\left| {\mathop {{\text{OB}}}\limits^ \to } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2}} = \left( {\sqrt {36} } \right)$

$\left| {\mathop {{\text{CD}}}\limits^ \to } \right| = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( { - 8} \right)}^2} + {{\left( 5 \right)}^2}} = \left( {\sqrt {125} } \right)$

correct substitution into $\frac{1}{2}bh$ A1

eg $\frac{1}{2} \times 6 \times \sqrt {125}$

area $= 3\sqrt {125} ,\,\,15\sqrt 5$ A1 N3

METHOD 2 (subtracting triangles)

recognizing that OB is altitude of either ΔOBD or ΔOBC(seen anywhere) M1

eg $\frac{1}{2} \times \left| {\mathop {{\text{OB}}}\limits^ \to } \right| \times \left| {\mathop {{\text{BD}}}\limits^ \to } \right|,\,\,{\text{OB}} \bot {\text{BC}},$ sketch of triangle showing right angle at B

one correct vector $\mathop {{\text{BD}}}\limits^ \to$ or $\mathop {{\text{DB}}}\limits^ \to$ or $\mathop {{\text{BC}}}\limits^ \to$ or $\mathop {{\text{CB}}}\limits^ \to$ (seen anywhere) (A1)

eg $\mathop {{\text{BD}}}\limits^ \to = \left( \begin{gathered} \,6 \hfill \\ \,8 \hfill \\ - 5 \hfill \\ \end{gathered} \right)$ , $\mathop {{\text{CB}}}\limits^ \to = \left( \begin{gathered} - 12 \hfill \\ - 16 \hfill \\ \,10 \hfill \\ \end{gathered} \right)$

$\left| {\mathop {{\text{OB}}}\limits^ \to } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2}} = \left( {\sqrt {36} } \right)$ (seen anywhere) (A1)

one correct magnitude of a base (seen anywhere) (A1)

$\left| {\mathop {{\text{BD}}}\limits^ \to } \right| = \sqrt {{{\left( 6 \right)}^2} + {{\left( 8 \right)}^2} + {{\left( 5 \right)}^2}} = \left( {\sqrt {125} } \right),\,\,\left| {\mathop {{\text{BC}}}\limits^ \to } \right| = \sqrt {144 + 256 + 100} = \left( {\sqrt {500} } \right)$

correct working A1

eg $\frac{1}{2} \times 6 \times \sqrt {500} - \frac{1}{2} \times 6 \times 5\sqrt 5 ,\,\,\frac{1}{2} \times 6 \times \sqrt {500} \times {\text{sin}}90 - \frac{1}{2} \times 6 \times 5\sqrt 5 \times {\text{sin}}90$

area $= 3\sqrt {125} ,\,\,15\sqrt 5$ A1 N3

METHOD 3 (using $\frac{1}{2}$ ab sin C with ΔOCD)

two correct side lengths (seen anywhere) (A1)(A1)

$\left| {\mathop {{\text{OD}}}\limits^ \to } \right| = \sqrt {{{\left( 8 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( { - 9} \right)}^2}} = \left( {\sqrt {161} } \right),\,\,\left| {\mathop {{\text{CD}}}\limits^ \to } \right| = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( { - 8} \right)}^2} + {{\left( 5 \right)}^2}} = \left( {\sqrt {125} } \right),\,$ $\left| {\mathop {{\text{OC}}}\limits^ \to } \right| = \sqrt {{{\left( {14} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 14} \right)}^2}} = \left( {\sqrt {536} } \right)$

attempt to find cosine ratio (seen anywhere) M1
eg $\frac{{536 - 286}}{{ - 2\sqrt {161} \sqrt {125} }},\,\,\frac{{{\text{OD}} \bullet {\text{DC}}}}{{\left| {OD} \right|\left| {DC} \right|}}$