2020HSCShort answer

Q13 (b)

3 marks

Question

Consider two lines in three dimensions given by:
$\mathbf r=\begin{pmatrix}3\\-1\\7\end{pmatrix}+\lambda_1\begin{pmatrix}1\\2\\1\end{pmatrix} \quad \text{and} \quad \mathbf r=\begin{pmatrix}3\\-6\\2\end{pmatrix}+\lambda_2\begin{pmatrix}-2\\1\\3\end{pmatrix}.$
By equating components, find the point of intersection of the two lines.

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	3
Correctly finds either $\lambda_1$ or $\lambda_2$ , or equivalent merit	2
Finds one equation linking $\lambda_1$ and $\lambda_2$ , or equivalent merit	1

Sample answer:

\vec r=\begin{pmatrix}3\\-1\\7\end{pmatrix}+\lambda_1\begin{pmatrix}1\\2\\1\end{pmatrix}

\vec r=\begin{pmatrix}3\\-6\\2\end{pmatrix}+\lambda_2\begin{pmatrix}-2\\1\\3\end{pmatrix}

Equating components,

3+\lambda_1=3-2\lambda_2\tag{1}

-1+2\lambda_1=-6+\lambda_2\tag{2}

7+\lambda_1=2+3\lambda_2\tag{3}

(1)\Rightarrow\lambda_1=-2\lambda_2

(2)\Rightarrow-1-4\lambda_2=-6+\lambda_2\Rightarrow\lambda_2=1\Rightarrow\lambda_1=-2

As a check, using equation (3), left hand side $=5$ , right hand side $=5$ .

\therefore\text{Point of intersection is }\vec r=\begin{pmatrix}3\\-6\\2\end{pmatrix}+\begin{pmatrix}2\\-1\\-3\end{pmatrix}=\begin{pmatrix}1\\-5\\5\end{pmatrix}

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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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