Science:Math Exam Resources/Courses/MATH152/April 2013/Question B 02 (c)

MATH152 April 2013

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Question B 02 (c)
Let $A=(0,1,2)$, $B=(1,0,3)$ and $C=(1,1,4)$. Let $\Delta$ be the triangle whose vertices are A,B, and C. Find the plane P containing $\Delta$ in the equation form i.e. in the form $\displaystyle ax+by+cz=d$ with a,b,c and d specified.

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Hint
If a point $\displaystyle X=(x_{0},y_{0},z_{0})$ lies in the plane $P$, then $\displaystyle ax_{0}+by_{0}+cz_{0}=d$.

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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. Since the points A, B, and C all lie in the plane P, we have the following equations are satisfied: ${\begin{aligned}b+2c&=d\\a+3c&=d\\a+b+4c&=d\end{aligned}}$. Writing these equations as a linear system gives ${\begin{bmatrix}0&1&2&-1\\1&0&3&-1\\1&1&4&-1\\\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}=\mathbf {0}$ Row reducing this system such that we have a pivot in each row we get: ${\begin{bmatrix}1&0&0&2\\0&1&0&1\\0&0&1&-1\\\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}=\mathbf {0} \quad \rightarrow \quad a=-2d,\,b=-d,\,c=d$ So we can see that we have one free parameter. In other words, any specific values of a,b, and c are determined by our choice of d. If we choose d = 1, then a = -2, b = -1, c = 1. Hence, the plane defined by the points A,B and C is ${\color {blue}-2x-y+z=1}.$ (Note that any constant multiple the planar equation is also a correct answer: e.g. $\displaystyle 4x+2y-2z=-2.$ )

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