Science:Math Exam Resources/Courses/MATH200/December 2012/Question 01 (b)/Solution 1

We are searching for a starting point $\displaystyle p$ and a direction vector $\displaystyle v$ for a line $\displaystyle M=p+\lambda v$.

Since, the point $\displaystyle {\begin{pmatrix}10\\11\\13\end{pmatrix}}$ lies on the line, we choose $\displaystyle p={\begin{pmatrix}10\\11\\13\end{pmatrix}}$

The direction vector $\displaystyle v$ must be perpendicular to the direction vector of $\displaystyle L$, such that $\displaystyle M$ and $\displaystyle L$ are perpendicular. Hence $\displaystyle v\circ {\begin{pmatrix}-3\\1\\2\end{pmatrix}}=-3v_{1}+v_{2}+2v_{3}=0$

The direction vector $\displaystyle v$ must lie in the plane $\displaystyle y=z$, then the line $\displaystyle M$ is parallel to the plane. Hence $\displaystyle v_{2}=v_{3}$.

Together, we obtain $\displaystyle -3v_{1}+v_{2}+2v_{2}=-3v_{1}+3v_{2}=0\quad \Rightarrow \quad v_{1}=v_{2}$

With $\displaystyle v_{1}=v_{2}=v_{3}$, we choose the vector $\displaystyle v$ to be $\displaystyle v={\begin{pmatrix}1\\1\\1\end{pmatrix}}$

and so $\displaystyle \ \ M={\begin{pmatrix}10\\11\\13\end{pmatrix}}+\lambda {\begin{pmatrix}1\\1\\1\end{pmatrix}}$