Science:Math Exam Resources/Courses/MATH105/April 2012/Question 08 (i)

MATH105 April 2012

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Question 08 (i)
Short-answer question! No credit will be given for the answer (even if it is correct) without the accompanying work. Find the equation of the plane parallel to 3x-y+4z=13 passing through the point (2,1,-1).

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
What is a normal vector to the given plane? What is a normal vector of the plane you are looking for? Given a normal vector and a point, finding the equation of the plane follows a simple formula.

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 a plane parallel to $3x-y+4z=13$, we can take its normal vector to be $<3,-1,4>$ (two parallel planes have normal vectors that are scalar multiples of each other and we can just as well choose their normal vectors to be identical). Thus, the plane we seek has equation $3(x-x_{0})+(-1)(y-y_{0})+4(z-z_{0})=0$ where $(x_{0},y_{0},z_{0})$ is any point on the plane. We are told the plane passes through $(2,1,-1)$ so we take $x_{0}=2$, $y_{0}=1$, and $z_{0}=-1$ and the plane's equation is: $3(x-2)-(y-1)+4(z-(-1))=0$ or equivalently, $3x-y+4z=1$.

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