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

MATH152 April 2022

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[hide] Question A11
Let ${\textstyle W_{1},\ W_{2}}$ be planes in 3-space ${\textstyle \mathbb {R} ^{3}}$ defined by equations $${\begin{aligned}W_{1}:&\ x_{1}-2x_{2}+x_{3}=1,\\W_{2}:&\ x_{1}-x_{2}+2x_{3}=2.\end{aligned}}$$ Find the parametric form of the line of intersection of these planes.

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[show] Hint
Recall that the parametric form of the line is ${\textstyle {\vec {x}}=p+t{\vec {v}}}$. To solve this problem you need to find the point ${\textstyle p}$ and the vector ${\textstyle {\vec {v}}}$. Can you use what you know about the normal vectors of planes to determine the direction vector ${\textstyle {\vec {v}}}$ of the line of intersection?

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[show] Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $$\left[{\begin{array}{c}x_{1}(t)\\x_{2}(t)\\x_{3}(t)\end{array}}\right]=\left[{\begin{array}{c}3\\1\\0\end{array}}\right]+t\left[{\begin{array}{c}-3\\-1\\1\end{array}}\right]$$

[show] Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Alternatively, this problem can be solved simply by solving the system equations we solved to find the point ${\textstyle p}$ in the previous solution. To recap, that system is $$\left[{\begin{array}{ccc|c}1&-2&1&1\\1&-1&2&2\end{array}}\right].$$ Because this is a system with more variables than equations, it will either have no solution, or infinitely many solutions. Looking at the normal vectors for the planes, we can see that they are not parallel, meaning that the planes will intersect along a line! Recall that the system simplified to $${\begin{aligned}&x_{1}=3(1-x_{3})\\&x_{2}=1-x_{3}\end{aligned}}.$$ Now, since ${\textstyle x_{3}}$ is a free variable, let ${\textstyle x_{3}=t}$, giving the system $$\left\lbrace {\begin{aligned}&x_{1}=3(1-t)\\&x_{2}=1-t\\&x_{3}=t\end{aligned}}\right.\rightarrow \left[{\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}}\right]=\left[{\begin{array}{c}3\\1\\0\end{array}}\right]+t\left[{\begin{array}{c}-3\\-1\\1\end{array}}\right].$$