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

MATH152 April 2012

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q6 (d)  •  Q6 (e)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q7 (d)  •  Q8 (a)  •  Q8 (b)  •  Q8 (c)  •  Q8 (d)  •  

   Other MATH152 Exams

  •  April 2016  •  April 2015  •  April 2014  •  April 2022  •  April 2011  •  April 2010  •  April 2012  •  April 2013  •  April 2017  •  

Question 01 (a)
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 the planes $S_{1}$ and $S_{2}$ in equation form.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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
Recall that in order to define a plane, we need a normal vector and a point on the plane.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. For the first plane $S_{1}$ we know that its normal is $\mathbf {n} =[2,1,-1]$ as this is provided. We’re also told that the point $P=[1,-1,0]$ is on the plane. Using this information we can easily get the equation for $S_{1}$ via ${\begin{aligned}0&=\mathbf {n} \cdot \left(\mathbf {x} -\mathbf {P} \right)\\&=[2,1,-1]\cdot [x_{1}-1,x_{2}+1,x_{3}]\\&=2x_{1}+x_{2}-x_{3}-1.\end{aligned}}$ We have to work a little harder to get the equation for the plane of $S_{2}$. However, notice we are given three points that are on the plane. Let $\mathbf {u} ={\vec {AB}}$ be the vector connecting A to B and $\mathbf {v} ={\vec {AC}}$ be the vector connecting A to C. We therefore have that, ${\begin{aligned}\mathbf {u} =B-A&=[-2,2,1]\\\mathbf {v} =C-A&=[-1,1,1].\end{aligned}}$ We know that the normal to the plane is orthogonal to all vectors on the plane. Since $\mathbf {u}$ and $\mathbf {v}$ are on the plane, if we take their cross product, we will get a vector orthogonal to both $\mathbf {u}$ and $\mathbf {v}$ and thus be orthogonal to the plane. Call this vector $\mathbf {m}$, ${\begin{aligned}\mathbf {m} &=\mathbf {u} \times \mathbf {v} \\&=[-2,2,1]\times [-1,1,1]\\&={\textrm {det}}{\begin{pmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\-2&2&1\\-1&1&1\end{pmatrix}}\\&=[1,1,0].\end{aligned}}$ We can now use this normal vector along with any point on the plane to get the equation. We will use $\mathbf {a} =[1,0,0]$, ${\begin{aligned}0=\mathbf {m} \cdot (\mathbf {x} -\mathbf {a} )=[1,1,0]\cdot [x_{1}-1,x_{2},x_{3}]=x_{1}+x_{2}-1.\end{aligned}}$ Therefore, we get that the equation for $S_{1}$ is $2x_{1}+x_{2}-x_{3}-1=0$ while the equation for $S_{2}$ is $x_{1}+x_{2}-1=0$.