Science:Math Exam Resources/Courses/MATH105/April 2013/Question 01 (a)

MATH105 April 2013

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Question 01 (a)
Short-Answer Questions. Put your answer in the box provided but show you work also. Each question is worth 3 marks, but not all questions are of equal difficulty. Find an equation of the plane which is parallel to the plane $\displaystyle Q:-2x+y=3z+1$ and passes through the point $\displaystyle (-1,1,2)$.

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Hint
Equations of parallel planes share the same normal vector $\displaystyle \langle n_{1},n_{2},n_{3}\rangle$, with $\displaystyle xn_{1}+yn_{2}+zn_{3}=a$

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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 we have the normal form of the plane, we find the vector, normal to the plane, in the equation. ${\begin{aligned}-2x+y=3z+1\qquad {\text{which is}}\qquad (x,y,z)\cdot \underbrace {\begin{pmatrix}-2\\1\\-3\end{pmatrix}} _{{\text{normal vector}}n}=1\end{aligned}}$ All equations of planes, parallel to the plane $\ Q$, have the same normal vector $\ n$. This means the equation looks like $(x,y,z)\cdot {\begin{pmatrix}-2\\1\\-3\end{pmatrix}}=a\in \mathbb {R}$ With the point $\displaystyle (-1,1,2)$ lying in the plane, we find $\displaystyle a$ by $a=(-1,1,2)\cdot {\begin{pmatrix}-2\\1\\-3\end{pmatrix}}=-3$ This gives us the equation of the plane ${\begin{aligned}(x,y,z)\cdot {\begin{pmatrix}-2\\1\\-3\end{pmatrix}}=-3\qquad {\text{which is}}\qquad -2x+y-3z=-3\end{aligned}}$

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