IBDP Math AI:Topic:AHL 3.11:Vector equation of a line in two and three dimensions-IB style Questions HL Paper 1

Ans:

(a) (one vector to the line is \(\binom{1}{m}\) therefore) a= (\(\binom{0}{c} \)
the line goes m up for every 1 across
(so the direction vector is) \(b= ( \binom{1}{m}\)
(b) (from GDC OR \(6 \times 2 – 4 \times 3) \) |M| = 0
(c) METHOD 1
\(\binom{X}{Y} = \begin{pmatrix}
6 & 3\\
4 & 2
\end{pmatrix} \binom{x}{mx+c}= \binom{6x + 3mx + 3c}{4x + 2mx + 2c}\)
therefore the new line has equation 3Y = 2X
which is independent of m or c
METHOD 2
take two points on the line, e.g. (0, c) and (1, m+c)
these map to \(\begin{pmatrix}
6 & 3\\
4 & 2
\end{pmatrix}\binom{0}{c}=\binom{3c}{2c}\)
and \(\begin{pmatrix}
6 & 3\\
4 & 2
\end{pmatrix}\binom{1}{m+c}=\binom{6+3m+3c}{4+2m+2c}\)
therefore a direction vector is \(\binom{6+3m}{4+2m}=(2+m)\binom{3}{2}\)
(since \(\neq -2\)) a direction vector is \(\binom{3}{2}\)
the line passes through \(\binom{3c}{2c})-c \biniom{3}{2} = \binom{0}{0}\) therefore it always has the
origin as a jump-on vector
the vector equation is therefore \(r=\mu \binom{3}{2}\)
which is independent of m or c
METHOD 3
\(r=\begin{pmatrix}
6 & 3\\
4 & 2
\end{pmatrix}(\binom{0}{c}+ \lambda \binom{1}{m})=\binom{3c}{2c}+ \lambda \binom{6+3m}{4+2m}\)
\(=c \binom{3}{2}+(2+m)\lambda \binom{3}{2}\)
\(=\mu \binom{3}{2}\)
where \(\mu=c+(2+m) \lambda\) is an arbitrary parameter.
which is independent of m or c (as \(\mu\) can take any value)