Science:Math Exam Resources/Courses/MATH152/April 2022/Question A05

MATH152 April 2022

  •  QA01  •  QA02  •  QA03  •  QA04  •  QA05  •  QA06  •  QA07  •  QA08  •  QA09  •  QA10  •  QA11  •  QA12  •  QA13  •  QA14  •  QA15  •  QA16  •  QA17  •  QA18  •  QA19  •  QA20  •  QB1 (a)  •  QB1 (b)  •  QB2 (a)  •  QB2 (b)  •  QB2 (c)  •  QB3 (a)  •  QB3 (b)  •  QB4 (a)  •  QB4 (b)  •  QB4 (c)  •  QB5 (a)  •  QB5 (b)  •  QB5 (c)  •  

   Other MATH152 Exams

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

Question A05
Find the volume of the parallelepiped with sides ${\vec {u}},{\vec {v}},{\vec {u}}\times {\vec {v}}$, where $${\vec {u}}=\left[{\begin{array}{c}1\\0\\1\end{array}}\right],\qquad {\vec {v}}=\left[{\begin{array}{c}1\\2\\0\end{array}}\right].$$

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
First, you need to compute ${\vec {u}}\times {\vec {v}}$. One way to do this is via the mnemonic $${\vec {u}}\times {\vec {v}}=\det \left[{\begin{array}{ccc}{\vec {i}}&{\vec {j}}&{\vec {k}}\\u_{1}&u_{2}&u_{3}\\v_{1}&v_{2}&v_{3}\end{array}}\right].$$

Hint 2
Second, recall that the cross-product of two vectors is orthogonal to both vectors and its norm is the area of the parallelogram spanned by the two vectors. Try sketching these vectors.

Hint 3
For a different approach, you might remember that the volume of a parallelepiped that is spanned by the 3-dimensional vectors ${\vec {u}},{\vec {v}},{\vec {w}}$ is equal to $$\det \left[{\begin{array}{ccc}u_{1}&u_{2}&u_{3}\\v_{1}&v_{2}&v_{3}\\w_{1}&w_{2}&w_{3}\end{array}}\right].$$

Checking a solution serves two purposes: helping you if, after having used all the hints, 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 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. By the first hint, we have $${\vec {u}}\times {\vec {v}}=\det \left[{\begin{array}{ccc}{\vec {i}}&{\vec {j}}&{\vec {k}}\\1&0&1\\1&2&0\end{array}}\right]=-2{\vec {i}}+{\vec {j}}+2{\vec {k}}=\left[{\begin{array}{c}-2\\1\\2\end{array}}\right].$$ By the second hint, the parallelepiped in question is a prism whose base has area $\|{\vec {u}}\times {\vec {v}}\|$, and whose height has length $\|{\vec {u}}\times {\vec {v}}\|$, so its volume is $$\|{\vec {u}}\times {\vec {v}}\|^{2}=9.$$

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. We provide the solution that follows the third hint. For that, we still need the computation ${\vec {u}}\times {\vec {v}}=-2{\vec {i}}+{\vec {j}}+2{\vec {k}}.$ By the second hint, the volume of the parallelepiped is $$\det \left[{\begin{array}{ccc}1&0&1\\1&2&0\\-2&1&2\end{array}}\right]=1\cdot \det \left[{\begin{array}{cc}2&0\\1&2\end{array}}\right]-0+1\cdot \det \left[{\begin{array}{cc}1&2\\-2&1\end{array}}\right]=9$$