2022TrialShort answer

Q15 (a) (ii)

3 marks

Question

Show by finding it that the point of intersection between the lines
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\10\\5\end{pmatrix}+\lambda\begin{pmatrix}2\\-3\\1\end{pmatrix}$
and
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}15\\-1\\6\end{pmatrix}+\delta\begin{pmatrix}4\\-1\\-1\end{pmatrix}$
is $\begin{pmatrix}7\\1\\8\end{pmatrix}$ .

Try the question before revealing the answer.

Solution

(ii) Point of intersection between

\begin{pmatrix}1\\10\\5\end{pmatrix}+\lambda\begin{pmatrix}2\\-3\\1\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}15\\-1\\6\end{pmatrix}+\delta\begin{pmatrix}4\\-1\\-1\end{pmatrix}.

x:\ 1+2\lambda=15+4\delta\Rightarrow\lambda-2\delta=7\quad(1)

y:\ 10-3\lambda=-1-\delta\Rightarrow3\lambda-\delta=11\quad(2)

z:\ 5+\lambda=6-\delta\Rightarrow\lambda+\delta=1\quad(3)

4\lambda=12

\lambda=3.

Sub. $\lambda=3$ in (3): $3+\delta=1$ .

\delta=-2.

Sub. $\lambda=3,\delta=-2$ in (1) to make sure it works:

3-2\times(-2)=7\ \checkmark

Point of intersection:

\begin{pmatrix}1\\10\\5\end{pmatrix}+3\begin{pmatrix}2\\-3\\1\end{pmatrix} \quad\text{or}\quad \begin{pmatrix}15\\-1\\6\end{pmatrix}-2\begin{pmatrix}4\\-1\\-1\end{pmatrix}

$=\begin{pmatrix}7\\1\\8\end{pmatrix}=\begin{pmatrix}7\\1\\8\end{pmatrix}.\checkmark$

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Vectors

Vectors and vector equations of linesUse coordinates and vector equations for curves, lines, line segments and spheres in two and three dimensions.

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