Science:Math Exam Resources/Courses/MATH152/April 2022/Question A11/Solution 1

The line of intersection of planes ${\textstyle W_{1}}$ and ${\textstyle W_{2}}$ will be perpendicular to both ${\textstyle n_{1}}$ (the normal vector of ${\textstyle W_{1}}$) and ${\textstyle n_{2}}$ (the normal vector of ${\textstyle W_{2}}$). Since the cross product of two linearly independent vectors in ${\textstyle \mathbb {R} ^{3}}$ results in a vector that is perpendicular to both of them, we will take the cross product of ${\textstyle n_{1}}$ and ${\textstyle n_{2}}$ to get the direction vector ${\textstyle {\vec {v}}}$ for the line of intersection.

$${\vec {v}}=n_{1}\times n_{2}=\det \left[{\begin{array}{ccc}i&j&k\\1&-2&1\\1&-1&2\end{array}}\right]=\left[{\begin{array}{c}-3\\-1\\1\end{array}}\right].$$

To find the point ${\textstyle p}$ we solve the system of equations

$$\left[{\begin{array}{ccc|c}1&-2&1&1\\1&-1&2&2\end{array}}\right]$$

using row operations.

$$\left[{\begin{array}{ccc|c}1&-2&1&1\\1&-1&2&2\end{array}}\right]\rightarrow \left[{\begin{array}{ccc|c}1&-2&1&1\\0&1&1&1\end{array}}\right]\rightarrow \left[{\begin{array}{ccc|c}1&0&3&3\\0&1&1&1\end{array}}\right].$$

We now have the system of equations

$${\begin{aligned}&x_{1}+3x_{3}=3\\&x_{2}+x_{3}=1\end{aligned}}$$

Rearranging and solving them for ${\textstyle x_{1}}$ and ${\textstyle x_{2}}$ gives: ${\textstyle x_{1}=3(1-x_{3})}$ and ${\textstyle x_{2}=1-x_{3}}$. Since ${\textstyle x_{3}}$ is a free variable now, we can choose ${\textstyle x_{3}=0}$ and substitute this into the expressions for ${\textstyle x_{1}}$ and ${\textstyle x_{2}}$ to determine the point ${\textstyle p}$. We have ${\textstyle p=(3,1,0)}$.

Concluding that the line is

$$\left[{\begin{array}{c}x_{1}(t)\\x_{2}(t)\\x_{3}(t)\end{array}}\right]=\left[{\begin{array}{c}3\\1\\0\end{array}}\right]+t\left[{\begin{array}{c}-3\\-1\\1\end{array}}\right]$$