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

MATH152 April 2012

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Question 01 (e)
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 $S_{2}$. Calculate the volume of the parallelpiped with edges AB, AC, and AP.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
How can we write the volume of a parallelepiped formed by three vectors in terms of a determinant?

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Solution

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 parallelepiped formed by the vectors $AB$, $AC$ and $AP$ is given by (the absolute value of) the determinant of the three vectors. From 1(a) and 1(c), we have that ${\begin{aligned}{\vec {AB}}&=[-2,2,1]\\{\vec {AC}}&=[-1,1,1]\end{aligned}}$ and so all we need is ${\vec {AP}}$, ${\vec {AP}}=P-A=[0,-1,0].$ Therefore, the volume of the parallelepiped is given by ${\begin{aligned}{\textrm {volume}}&=\left|{\textrm {det}}{\begin{pmatrix}-2&2&1\\-1&1&1\\0&-1&0\end{pmatrix}}\right|\\&=\left|-1{\textrm {det}}{\begin{pmatrix}-2&1\\-1&1\end{pmatrix}}\right|\\&=|-2+1|=|-1|=1.\end{aligned}}$