Science:Math Exam Resources/Courses/MATH221/December 2008/Question 04 (b)

MATH221 December 2008

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Question 04 (b)
Let W be the plane in $\mathbb {R} ^{4}$ spanned by the vectors $v={\begin{pmatrix}1\\1\\1\\1\end{pmatrix}}$ and $w={\begin{pmatrix}2\\0\\2\\0\end{pmatrix}}$. Find the vector in W which is closest to $b={\begin{pmatrix}1\\2\\3\\4\end{pmatrix}}$.

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
Think about orthogonal projections. Part (a) could be helpful.

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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. The vector in W which is closest to b will be the orthogonal projection of b onto W . From part (a), we know that the vectors $u={\begin{pmatrix}0\\1\\0\\1\end{pmatrix}}$ and $w={\begin{pmatrix}2\\0\\2\\0\end{pmatrix}}$ form an orthogonal basis for W. Let us make an orthonormal basis by normalizing u and w to ${\begin{aligned}e_{1}={\frac {u}{\|u\|}}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1\\0\\1\\0\end{pmatrix}}\quad {\text{and}}\quad e_{2}={\frac {w}{\|w\|}}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}0\\1\\0\\1\end{pmatrix}}.\end{aligned}}$ Then the orthogonal projection of b onto W is ${\begin{aligned}P_{W}(b)=\langle b,e_{1}\rangle e_{1}+\langle b,e_{2}\rangle e_{2}={\frac {1}{\sqrt {2}}}(1\cdot 1+3\cdot 1)e_{1}+{\frac {1}{\sqrt {2}}}(2\cdot 1+4\cdot 1)e_{2}=2{\sqrt {2}}e_{1}+3{\sqrt {2}}e_{2}={\begin{pmatrix}2\\3\\2\\3\end{pmatrix}}.\end{aligned}}$. Thus ${\begin{pmatrix}2\\3\\2\\3\end{pmatrix}}$ is the vector in W closest to the vector $b={\begin{pmatrix}1\\2\\3\\4\end{pmatrix}}.$

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