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Date November 2016 Marks available 3 Reference code 16N.1.sl.TZ0.4
Level SL only Paper 1 Time zone TZ0
Command term Find Question number 4 Adapted from N/A

Question

The position vectors of points P and Q are i \( + \) 2 j \( - \) k and 7i \( + \) 3j \( - \) 4k respectively.

Find a vector equation of the line that passes through P and Q.

[4]
a.

The line through P and Q is perpendicular to the vector 2i \( + \) nk. Find the value of \(n\).

[3]
b.

Markscheme

valid attempt to find direction vector     (M1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{PQ}}} ,{\text{ }}\overrightarrow {{\text{QP}}} \)

correct direction vector (or multiple of)     (A1)

eg\(\,\,\,\,\,\)6i \( + \) j \( - \) 3k

any correct equation in the form r \( = \) a \( + \) tb (any parameter for \(t\))     A2     N3

where a is i \( + \) 2j \( - \) k or 7i \( + \) 3j \( - \) 4k , and b is a scalar multiple of 6i \( + \) j \( - \) 3k

eg\(\,\,\,\,\,\)r \( = \) 7i \( + \) 3j \( - \) 4k \( + \) t(6i \( + \) j \( - \) 3k), r \( = \left( {\begin{array}{*{20}{c}} {1 + 6s} \\ {2 + 1s} \\ { - 1 - 3s} \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 6} \\ { - 1} \\ 3 \end{array}} \right)\)

 

Notes: Award A1 for the form a \( + \) tb, A1 for the form L \( = \) a \( + \) tb, A0 for the form r \( = \) b \( + \) ta.

 

[4 marks]

a.

correct expression for scalar product     (A1)

eg\(\,\,\,\,\,\)\(6 \times 2 + 1 \times 0 + ( - 3) \times n,{\text{ }} - 3n + 12\)

setting scalar product equal to zero (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)u \( \bullet \) v \( = 0,{\text{ }} - 3n + 12 = 0\)

\(n = 4\)    A1     N2

[3 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 4 - Vectors » 4.2 » The scalar product of two vectors.
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