IB DP Maths Topic 4.1 Algebraic and geometric approaches to unit vectors; base vectors; i, j and k. SL Paper 1

 

https://play.google.com/store/apps/details?id=com.iitianDownload IITian Academy  App for accessing Online Mock Tests and More..

Question

A particle is moving with a constant velocity along line L . Its initial position is A(6 , −2 , 10) . After one second the particle has moved to B( 9, −6 , 15) .

(i)     Find the velocity vector, \(\overrightarrow {{\rm{AB}}} \) .

(ii)    Find the speed of the particle.

[4]
a(i) and (ii).

Write down an equation of the line L .

[2]
b.
Answer/Explanation

Markscheme

(i) evidence of approach     (M1)

e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \({\rm{B}} – {\rm{A}}\) , \(\left( {\begin{array}{*{20}{c}}
{9 – 6}\\
{ – 6 + 2}\\
{15 – 10}
\end{array}} \right)\)

\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
3\\
{ – 4}\\
5
\end{array}} \right)\) (accept \(\left( {3, – 4, 5} \right)\) )    A1     N2

(ii) evidence of finding the magnitude of the velocity vector     M1

e.g. \({\text{speed}} = \sqrt {{3^2} + {4^2} + {5^2}} \)

\({\text{speed}} = \sqrt {50} \) \(\left( { = 5\sqrt 2 } \right)\)     A1     N1

[4 marks]

a(i) and (ii).

correct equation (accept Cartesian and parametric forms)     A2     N2

e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
{10}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
{ – 4}\\
5
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
9\\
{ – 6}\\
{15}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
{ – 4}\\
5
\end{array}} \right)\)

[2 marks]

b.

Question

The vertices of the triangle PQR are defined by the position vectors

\(\overrightarrow {{\rm{OP}}}  = \left( {\begin{array}{*{20}{c}}
4\\
{ – 3}\\
1
\end{array}} \right)\) , \(\overrightarrow {{\rm{OQ}}}  = \left( {\begin{array}{*{20}{c}}
3\\
{ – 1}\\
2
\end{array}} \right)\) and \(\overrightarrow {{\rm{OR}}}  = \left( {\begin{array}{*{20}{c}}
6\\
{ – 1}\\
5
\end{array}} \right)\) .

Find

(i)     \(\overrightarrow {{\rm{PQ}}} \) ;

(ii)    \(\overrightarrow {{\rm{PR}}} \) .

[3]
a.

Show that \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) .

[7]
b.

(i)     Find \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}}\) .

(ii)    Hence, find the area of triangle PQR, giving your answer in the form \(a\sqrt 3 \) .

[6]
c.
Answer/Explanation

Markscheme

(i) evidence of approach     (M1)

e.g. \(\overrightarrow {{\rm{PQ}}}  = \overrightarrow {{\rm{PO}}}  + \overrightarrow {{\rm{OQ}}} \) , \({\rm{Q}} – {\rm{P}}\)

\(\overrightarrow {{\rm{PQ}}}  = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
1
\end{array}} \right)\)     A1     N2

(ii) \(\overrightarrow {{\rm{PR}}}  = \left( {\begin{array}{*{20}{c}}
2\\
2\\
4
\end{array}} \right)\)     A1     N1

[3 marks]

a.

METHOD 1

choosing correct vectors \(\overrightarrow {{\rm{PQ}}} \) and \(\overrightarrow {{\rm{PR}}} \)    (A1)(A1)

finding \(\overrightarrow {{\rm{PQ}}}  \bullet \overrightarrow {{\rm{PR}}} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right|\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right|\)    (A1) (A1)(A1)

\(\overrightarrow {{\rm{PQ}}}  \bullet \overrightarrow {{\rm{PR}}}  = – 2 + 4 + 4( = 6)\)

\(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt {{{( – 1)}^2} + {2^2} + {1^2}} \) \(\left( { = \sqrt 6 } \right)\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {{2^2} + {2^2} + {4^2}} \) \(\left( { = \sqrt {24} } \right)\)

substituting into formula for angle between two vectors     M1

e.g. \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{6}{{\sqrt 6  \times \sqrt {24} }}\)

simplifying to expression clearly leading to \(\frac{1}{2}\)     A1

e.g. \(\frac{6}{{\sqrt 6  \times 2\sqrt 6 }}\) , \(\frac{6}{{\sqrt {144} }}\) , \(\frac{6}{{12}}\)

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\)     AG     N0

METHOD 2

evidence of choosing cosine rule (seen anywhere)     (M1)

\(\overrightarrow {{\rm{QR}}}  = \left( {\begin{array}{*{20}{c}}
3\\
0\\
3
\end{array}} \right)\)     A1

\(\left| {\overrightarrow {{\rm{QR}}} } \right| = \sqrt {18} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt 6 \) and \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {24} \)     (A1)(A1)(A1)

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt {24} } \right)}^2} – {{\left( {\sqrt {18} } \right)}^2}}}{{2\sqrt 6  \times \sqrt {24} }}\)     A1

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{6 + 24 – 18}}{{24}}\) \(\left( { = \frac{{12}}{{24}}} \right)\)     A1

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\)      AG     N0

[7 marks]

b.

(i) METHOD 1

evidence of appropriate approach     (M1)

e.g. using \({\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q + co}}{{\rm{s}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q}} = 1\) , diagram

substituting correctly     (A1)

e.g. \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {1 – {{\left( {\frac{1}{2}} \right)}^2}} \)

\({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {\frac{3}{4}} \) \(\left( { = \frac{{\sqrt 3 }}{2}} \right)\)     A1     N3

METHOD 2

since \(\cos \widehat {\rm{P}} = \frac{1}{2}\) , \(\widehat {\rm{P}} = 60^\circ \)     (A1)

evidence of approach

e.g. drawing a right triangle, finding the missing side     (A1)

\(\sin \widehat {\rm{P}} = \frac{{\sqrt 3 }}{2}\)     A1     N3

(ii) evidence of appropriate approach      (M1)

e.g. attempt to substitute into \(\frac{1}{2}ab\sin C\)

correct substitution

e.g. area \( = \frac{1}{2}\sqrt 6  \times \sqrt {24}  \times \frac{{\sqrt 3 }}{2}\)     A1

area \( = 3\sqrt 3 \)     A1     N2

[6 marks]

c.

Question

The line \({L_1}\) is parallel to the z-axis. The point P has position vector \(\left( {\begin{array}{*{20}{c}}
8\\
1\\
0
\end{array}} \right)\) and lies on \({L_1}\).

Write down the equation of \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\).

[2]
a.

The line \({L_2}\) has equation \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
2\\
4\\
{ – 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right)\) . The point A has position vector \(\left( {\begin{array}{*{20}{c}}
6\\
2\\
9
\end{array}} \right)\) .

Show that A lies on \({L_2}\) .

[4]
b.

Let B be the point of intersection of lines \({L_1}\) and \({L_2}\) .

(i)     Show that \(\overrightarrow {{\rm{OB}}}  = \left( {\begin{array}{*{20}{c}}
8\\
1\\
{14}
\end{array}} \right)\) .

(ii)    Find \(\overrightarrow {{\rm{AB}}} \) .

[7]
c.

The point C is at (2, 1, − 4). Let D be the point such that ABCD is a parallelogram.

Find \(\overrightarrow {{\rm{OD}}} \) .

[3]
d.
Answer/Explanation

Markscheme

\({L_1}:{\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
8\\
1\\
0
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
0\\
0\\
1
\end{array}} \right)\)     A2     N2

[2 marks]

a.

evidence of equating \({\boldsymbol{r}}\) and \(\overrightarrow {{\rm{OA}}} \)     (M1)

e.g. \(\left( {\begin{array}{*{20}{c}}
6\\
2\\
9
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2\\
4\\
{ – 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right)\) , \(A = r\)

one correct equation     A1

e.g. \(6 = 2 + 2s\) , \(2 = 4 – s\) , \(9 = – 1 + 5s\)

\(s = 2\)     A1

evidence of confirming for other two equations     A1

e.g. \(6 = 2 + 4\) , \(2 = 4 – 2\) , \(9 = – 1 + 10\)

so A lies on \({L_2}\)      AG     N0

[4 marks]

b.

(i) evidence of approach     M1

e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
4\\
{ – 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
8\\
1\\
0
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
0\\
0\\
1
\end{array}} \right)\) \({L_1} = {L_2}\)

one correct equation     A1

e.g. \(2 + 2s = 8\) , \(4 – s = 1\) , \( – 1 + 5s = t\)

attempt to solve     (M1)

finding \(s = 3\)     A1

substituting     M1

e.g. \(\overrightarrow {{\rm{OB}}}  = \left( {\begin{array}{*{20}{c}}
2\\
4\\
{ – 1}
\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right)\)

\(\overrightarrow {{\rm{OB}}}  = \left( {\begin{array}{*{20}{c}}
8\\
1\\
{14}
\end{array}} \right)\)     AG     N0

(ii) evidence of appropriate approach     (M1)

e.g. \(\overrightarrow {{\rm{AB}}}  = \overrightarrow {{\rm{AO}}}  + \overrightarrow {{\rm{OB}}} \) , \(\overrightarrow {{\rm{AB}}}  = \overrightarrow {{\rm{OB}}}  – \overrightarrow {{\rm{OA}}} \)

\(\overrightarrow {{\rm{AB}}}  = \left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right)\)     A1     N2  

[7 marks]

c.

evidence of appropriate approach    (M1)

e.g. \(\overrightarrow {{\rm{AB}}}  = \overrightarrow {{\rm{DC}}} \)

correct values     A1

e.g. \(\overrightarrow {{\rm{OD}}}  + \left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) ,  \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{2 – x}\\
{1 – y}\\
{ – 4 – z}
\end{array}} \right)\)

\(\overrightarrow {{\rm{OD}}}  = \left( {\begin{array}{*{20}{c}}
0\\
2\\
{ – 9}
\end{array}} \right)\)     A1     N2

[3 marks]

d.

Question

Let \(\overrightarrow {{\rm{AB}}}  = \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
3
\end{array}} \right)\) and \(\overrightarrow {{\rm{AC}}}  = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
{ – 3}\\
2
\end{array}} \right)\) .

Find \(\overrightarrow {{\rm{BC}}} \) .

[2]
a.

Find a unit vector in the direction of \(\overrightarrow {{\rm{AB}}} \) .

[3]
b.

Show that \(\overrightarrow {{\rm{AB}}} \) is perpendicular to \(\overrightarrow {{\rm{AC}}} \) .

[3]
c.
Answer/Explanation

Markscheme

evidence of appropriate approach     (M1)

e.g. \(\overrightarrow {{\rm{BC}}} {\rm{ = }}\overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{AC}}} \) , \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
{ – 3}\\
2
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
3
\end{array}} \right)\)

\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}
{ – 8}\\
{ – 1}\\
{ – 1}
\end{array}} \right)\)     A1     N2

[2 marks]

a.

attempt to find the length of \(\overrightarrow {{\rm{AB}}} \)     (M1)

\(\overrightarrow {|{\rm{AB}}} | = \sqrt {{6^2} + {{( – 2)}^2} + {3^2}} \) \(( = \sqrt {36 + 4 + 9} = \sqrt {49} = 7)\)     (A1)

unit vector is \(\frac{1}{7}\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
3
\end{array}} \right)\) \(\left( { = \left( {\begin{array}{*{20}{c}}
{\frac{6}{7}}\\
{ – \frac{2}{7}}\\
{\frac{3}{7}}
\end{array}} \right)} \right)\)     A1     N2

[3 marks]

b.

recognizing that the dot product or \(\cos \theta \) being 0 implies perpendicular     (M1)

correct substitution in a scalar product formula     A1

e.g. \((6) \times ( – 2) + ( – 2) \times ( – 3) + (3) \times (2)\) , \(\cos \theta  = \frac{{ – 12 + 6 + 6}}{{7 \times \sqrt {17} }}\)

correct calculation     A1

e.g. \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AC}}} = 0\) , \(\cos \theta = 0\)

therefore, they are perpendicular     AG     N0

[3 marks]

c.

Question

In the following diagram, \(\overrightarrow {{\text{OP}}} \) = p, \(\overrightarrow {{\text{OQ}}} \) = q and \(\overrightarrow {{\text{PT}}}  = \frac{1}{2}\overrightarrow {{\text{PQ}}} \).

Express each of the following vectors in terms of p and q,

\(\overrightarrow {{\text{QP}}} \);

[2]
a.

\(\overrightarrow {{\text{OT}}} \).

[3]
b.
Answer/Explanation

Markscheme

appropriate approach     (M1)

eg     \(\overrightarrow {{\text{QP}}}  = \overrightarrow {{\text{QO}}}  + \overrightarrow {{\text{OP}}} ,{\text{ P}} – {\text{Q}}\)

\(\overrightarrow {{\text{QP}}} \) = pq     A1     N2

[2 marks]

a.

recognizing correct vector for \(\overrightarrow {{\text{QT}}} \) or \(\overrightarrow {{\text{PT}}} \)     (A1)

eg     \(\overrightarrow {{\text{QT}}}  = \frac{1}{2}\)(pq), \(\overrightarrow {{\text{PT}}}  = \frac{1}{2}\)(qp)

appropriate approach     (M1)

eg     \(\overrightarrow {{\text{OT}}}  = \overrightarrow {{\text{OP}}}  + \overrightarrow {{\text{PT}}} ,{\text{ }}\overrightarrow {{\text{OQ}}}  + \overrightarrow {{\text{QT}}} ,{\text{ }}\overrightarrow {{\text{OP}}}  + \frac{1}{2}\overrightarrow {{\text{PQ}}} \)

\(\overrightarrow {{\text{OT}}}  = \frac{1}{2}\)(p + q)   \(\left( {{\text{accept }}\frac{{p + q}}{2}} \right)\)     A1     N2

[3 marks]

b.

Question

The line \({L_1}\) passes through the points \(\rm{A}(2, 1, 4)\) and \(\rm{B}(1, 1, 5)\).

Another line \({L_2}\) has equation r = \(\left( \begin{array}{c}4\\7\\ – 4\end{array} \right) + s\left( \begin{array}{c}0\\ – 1\\1\end{array} \right)\) . The lines \({L_1}\) and \({L_2}\) intersect at the point P.

Show that \(\overrightarrow {{\text{AB}}}  = \)  \(\left( \begin{array}{c} – 1\\0\\1\end{array} \right)\)

[1]
a.

Hence, write down a direction vector for \({L_1}\);

[1]
b(i).

Hence, write down a vector equation for \({L_1}\).

[2]
b(ii).

Find the coordinates of P.

[6]
c.

Write down a direction vector for \({L_2}\).

[1]
d(i).

Hence, find the angle between \({L_1}\) and \({L_2}\).

[6]
d(ii).
Answer/Explanation

Markscheme

correct approach     A1

eg   \(\left( \begin{array}{c}1\\1\\5\end{array} \right) – \left( \begin{array}{l}2\\1\\4\end{array} \right)\), \({\text{AO}} + {\text{OB}}\), \(b – a\)

\(\overrightarrow {{\text{AB}}}  = \)  \(\left( \begin{array}{c} – 1\\0\\1\end{array} \right)\)    AG     N0

[1 mark]

a.

correct vector (or any multiple)     A1     N1

eg     d =  \(\left( \begin{array}{c} – 1\\0\\1\end{array} \right)\)

[1 mark]

b(i).

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

where a is \(\left( \begin{array}{c}2\\1\\4\end{array} \right)\) or \(\left( \begin{array}{c}1\\1\\5\end{array} \right)\) , and b is a scalar multiple of  \(\left( \begin{array}{c} – 1\\0\\1\end{array} \right)\)     A2     N2

eg   r = \(\left( \begin{array}{c}1\\1\\5\end{array} \right) + t\left( \begin{array}{c} – 1\\0\\1\end{array} \right),\left( \begin{array}{c}x\\y\\z\end{array} \right) = \left( \begin{array}{c}2 – s\\1\\4 + s\end{array} \right)\)

 

Note:     Award A1 for a + tb, A1 for \({L_1}\) = a + tb, A0 for r = b + ta.

 

[2 marks]

b(ii).

valid approach     (M1)

eg     \({r_1} = {r_2}\), \(\left( \begin{array}{c}2\\1\\4\end{array} \right) + t\left( \begin{array}{c} – 1\\0\\1\end{array} \right) = \left( \begin{array}{c}4\\7\\ – 4\end{array} \right) + s\left( \begin{array}{c}0\\ – 1\\1\end{array} \right)\)

one correct equation in one parameter     A1

eg     \(2 – t = 4, 1 = 7 – s, 1 – t = 4\)

attempt to solve     (M1)

eg     \(2 – 4 = t, s = 7 – 1, t = 1 – 4\)

one correct parameter     A1

eg     \(t = -2, s = 6, t = -3\),

attempt to substitute their parameter into vector equation     (M1)

eg     \(\left( \begin{array}{c}4\\7\\ – 4\end{array} \right) + 6\left( \begin{array}{c}0\\ – 1\\1\end{array} \right)\)

P(4, 1, 2)   (accept position vector)     A1     N2

[6 marks]

c.

correct direction vector for \({L_2}\)     A1     N1

eg     \(\left( \begin{array}{c}0\\-1\\ 1\end{array} \right)\), \(\left( \begin{array}{c}0\\2\\ – 2\end{array} \right)\)

[1 mark]

d(i).

correct scalar product and magnitudes for their direction vectors     (A1)(A1)(A1)

scalar product \( = 0 \times  – 1 +  – 1 \times 0 + 1 \times 1{\text{ }}( = 1)\)

magnitudes \( = \sqrt {{0^2} + {{( – 1)}^2} + {1^2}} ,{\text{ }}\sqrt { – {1^2} + {0^2} + {1^2}} \left( {\sqrt 2 ,{\text{ }}\sqrt 2 } \right)\)

attempt to substitute their values into formula     M1

eg   \(\frac{{0 + 0 + 1}}{{\left( {\sqrt {{0^2} + {{( – 1)}^2} + {1^2}} } \right) \times \left( {\sqrt { – {1^2} + {0^2} + {1^2}} } \right)}},{\text{ }}\frac{1}{{\sqrt 2  \times \sqrt 2 }}\)

correct value for cosine, \(\frac{1}{2}\)     A1

angle is \(\frac{\pi }{3}{\text{ }}( = {60^ \circ })\)     A1     N1

[6 marks]

d(ii).
Scroll to Top