Science:Math Exam Resources/Courses/MATH152/April 2012/Question 01 (c)

MATH152 April 2012

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[hide] Question 01 (c)
Let $S_{1}$ be the plane perpendicular to $\mathbf {n} =[2,1,-1]$ and through the point P=[1,-1,0]. Let $S_{2}$ be the plane through the points A=[1,0,0], B=[-1,2,1], C=[0,1,1]. By L, we denote the line of intersection of $S_{1}$ and $S_{2}$. Determine L in parametric form.

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[show] Hint
We want to determine the line in terms of a single parameter. How does this relate to free variables in a linear system solution?

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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. From 1(b) we saw that the system of equations the line must satisfy is given by ${\begin{aligned}2x_{1}+x_{2}-x_{3}&=1\\x_{1}+x_{2}&=1.\end{aligned}}$ In order to find the parametric form we solve this system. First we put it in augmented form $\left[{\begin{array}{ccc|c}2&1&-1&1\\1&1&0&1\end{array}}\right]$ and then perform row operations to reduce the matrix. First we will swap row 1 and row 2 so that the first pivot (row 1, column 1) has value 1. $\left[{\begin{array}{ccc|c}1&1&0&1\\2&1&-1&1\end{array}}\right]$ and then we will subtract 2 multiples of row 1 from row 2 to place a zero below the pivot, $\left[{\begin{array}{ccc|c}1&1&0&1\\0&-1&-1&-1\end{array}}\right].$ We will then multiply the second row by -1 to put a 1 in the pivot position there (row 2, column 2), $\left[{\begin{array}{ccc|c}1&1&0&1\\0&1&1&1\end{array}}\right]$ and finally we will subtract 1 multiple of row 2 from row 1 to put a 0 above the pivot, $\left[{\begin{array}{ccc|c}1&0&-1&0\\0&1&1&1\end{array}}\right].$ Notice that because we have more columns than rows, then we have a free variables (the matrix is rank deficient). Let the free variable be $x_{3}$, i.e. let $x_{3}=t$. Then from our reduced matrix problem we have that $x_{1}=t$ and that $x_{2}+t=1$ or $x_{2}=1-t$. Therefore we can write that ${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}0\\1\\0\end{bmatrix}}+t{\begin{bmatrix}1\\-1\\1\end{bmatrix}}.$ This parametrically describes the line.

[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. From 1(a) we have that the normal directions to the two planes are ${\begin{aligned}\mathbf {n} &=[2,1,-1]\\\mathbf {m} &=[1,1,0]\end{aligned}}$ We know that the line must lie on both planes and therefore the direction of the line must be orthogonal to both normals. Therefore we can find it by taking the cross product, $\mathbf {n} \times \mathbf {m} =[2,1,-1]\times [1,1,0]={\textrm {det}}{\begin{pmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\2&1&-1\\1&1&0\end{pmatrix}}=[1,-1,1].$ We then must identify a point on the line. To do this we attempt to find one of the intercepts with the coordinate planes (one of $x_{1}$, $x_{2}$, or $x_{3}$ being zero). Set $x_{1}$ to be zero, from the plane equation for $S_{2}$, this tells us that $x_{2}=1$. Then using the plean equation for $S_{1}$, we get that $x_{3}=0$. Therefore, the point $(0,1,0)$ is on the line. This tells us that we can write the equation of the line as ${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}0\\1\\0\end{bmatrix}}+t{\begin{bmatrix}1\\-1\\1\end{bmatrix}}$ which is the parametric equation of the line.