Science:Math Exam Resources/Courses/MATH221/April 2013/Question 12 (b)

MATH221 April 2013

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Question 12 (b)
Consider the plane x - y + z = 0. b) Let u = $\left[{\begin{array}{c}1\\1\\1\end{array}}\right]$. Find vectors v and w such that u = v + w, where v is in the plane and w is perpendicular to the plane.

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 happens when you apply the projection of (a) to the vector u?

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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. If we apply the projection of (a) to the vector u, we get a vector w that is orthogonal to the plane. To get v, we simply subtract w from u. To verify that v is indeed on the plane, we compute that: $P_{1}(v)=P_{1}(u-w)=P_{1}(u)-P_{1}(w)=w-w=0$, which shows that the vector v has no competent that is orthogonal to the plane. Thus v must lie in the plane itself. Let's perform the necessary calculations to get w and v. We have $w={\frac {u\cdot \left[{\begin{array}{ccc}1&-1&1\\\end{array}}\right]}{3}}\left[{\begin{array}{ccc}1&-1&1\\\end{array}}\right]={\frac {1}{3}}\left[{\begin{array}{ccc}1&-1&1\\\end{array}}\right]$ And hence $v=u-w=\left[{\begin{array}{ccc}{\frac {2}{3}}&{\frac {4}{3}}&{\frac {2}{3}}\\\end{array}}\right]$. Note: It is also easy to directly verify that v satisfies the equation x - y + z = 0 of the plane, since $2/3-4/3+2/3=0$.

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