Science:Math Exam Resources/Courses/MATH221/April 2010/Question 02 (c)

MATH221 April 2010

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Question 02 (c)
Let $u_{1}={\begin{bmatrix}0\\2\\1\\2\end{bmatrix}}$, $u_{2}={\begin{bmatrix}1\\-2\\0\\2\end{bmatrix}}$, $u_{3}={\begin{bmatrix}2\\0\\2\\-1\end{bmatrix}}$ (c) Find a vector $u_{4}$ orthogonal to all of $u_{1},u_{2},u_{3}$.

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

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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. To find a vector orthogonal to all of $u_{1},u_{2},u_{3}$, we must find a vector whose dot product with each of $u_{1},u_{2},u_{3}$ is 0. Similar to part (b), we can arrange $u_{1},u_{2},u_{3}$ in a matrix. $M={\begin{bmatrix}u1&u2&u3\end{bmatrix}}={\begin{bmatrix}0&1&2\\2&-2&0\\1&0&2\\2&2&-1\end{bmatrix}}$ To find the vector we are looking for, we can solve for the null space of the transpose of $M$. This will satisfy the constraint that the dot product of the vector $u_{4}$ and each of $u_{1},u_{2},u_{3}$ is 0. So we solve: $M^{T}u_{4}={\begin{bmatrix}0\\0\\0\end{bmatrix}}$ This gives us: $u_{4}={\begin{bmatrix}-2/3\\-1/3\\2/3\\0\end{bmatrix}}$