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Date November 2012 Marks available 6 Reference code 12N.1.sl.TZ0.6
Level SL only Paper 1 Time zone TZ0
Command term Find Question number 6 Adapted from N/A

Question

The line L passes through the point (5,4,10) and is parallel to the vector (425) .

Write down a vector equation for line L .

[2]
a.

The line L intersects the x-axis at the point P. Find the x-coordinate of P.

[6]
b.

Markscheme

any correct equation in the form {\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}} (accept any parameter for t)

where a is \left( {\begin{array}{*{20}{c}} 5\\ { - 4}\\ {10} \end{array}} \right) , and b is a scalar multiple of \left( {\begin{array}{*{20}{c}} 4\\ { - 2}\\ 5 \end{array}} \right)     A2     N2

e.g. {\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}   5 \\   { - 4} \\   {10} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}}   4 \\   { - 2} \\   5 \end{array}} \right){\text{, }}{\boldsymbol{r}} = 5{\boldsymbol{i}} - 4{\boldsymbol{j}} + 10{\boldsymbol{k}} + t( - 8{\boldsymbol{i}} + 4{\boldsymbol{j}} - 10{\boldsymbol{k}})

Note: Award A1 for the form {\boldsymbol{a}} + t{\boldsymbol{b}} , A1 for L = {\boldsymbol{a}} + t{\boldsymbol{b}} , A0 for {\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}} .

 

[2 marks]

 

a.

recognizing that y = 0 or z = 0 at x-intercept (seen anywhere)     (R1) 

attempt to set up equation for x-intercept (must suggest x \ne 0 )     (M1)

e.g. L = \left( {\begin{array}{*{20}{c}} x\\ 0\\ 0 \end{array}} \right) 5 + 4t = xr = \left( {\begin{array}{*{20}{c}} 1\\ 0\\ 0 \end{array}} \right)

one correct equation in one variable     (A1)

e.g. - 4 - 2t = 0 10 + 5t = 0

finding t = - 2     A1 

correct working     (A1)

e.g. x = 5 + ( - 2)(4)

x = - 3 (accept ( - 3{\text{, }}0{\text{, }}0))     A1     N3

 

[6 marks]

 

b.

Examiners report

In part (a), the majority of candidates correctly recognized the equation that contains the position and direction vectors of a line. However, we saw a large number of candidates who continue to write their equations using "L = ", rather than the mathematically correct "{\boldsymbol{r}} = " or "\left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = ". r and \left( {\begin{array}{*{20}{c}}   x \\   y \\   z \end{array}} \right) represent vectors, whereas L is simply the name of the line. For part (b), very few candidates recognized that a general point on the x-axis will be given by the vector \left( {\begin{array}{*{20}{c}} x\\ 0\\ 0 \end{array}} \right) . Common errors included candidates setting their equation equal to\left( {\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right)  , or \left( {\begin{array}{*{20}{c}} 1\\ 0\\ 0 \end{array}} \right) , or even just the number 0.

a.

In part (a), the majority of candidates correctly recognized the equation that contains the position and direction vectors of a line. However, we saw a large number of candidates who continue to write their equations using "L = ", rather than the mathematically correct "{\boldsymbol{r}} = " or "\left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = ". r and \left( {\begin{array}{*{20}{c}}   x \\   y \\   z \end{array}} \right) represent vectors, whereas L is simply the name of the line. For part (b), very few candidates recognized that a general point on the x-axis will be given by the vector \left( {\begin{array}{*{20}{c}} x\\ 0\\ 0 \end{array}} \right) . Common errors included candidates setting their equation equal to\left( {\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right)  , or \left( {\begin{array}{*{20}{c}} 1\\ 0\\ 0 \end{array}} \right) , or even just the number 0.

b.

Syllabus sections

Topic 4 - Vectors » 4.3 » Vector equation of a line in two and three dimensions: r = a + tb .
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