Solution: Two vectors are orthogonal if their dot product is zero. Compute:
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- Solving equations 3 and 4: From 3, $ c = 7 - 3d $. Substituting into 4: $ 2(7 - 3d) + 4d = 8 \Rightarrow 14 - 6d + 4d = 8 \Rightarrow -2d = -6 \Rightarrow d = 3 $. Then $ c = 7 - 9 = -2 $.
- Thus, $ \mathbf{M} = \begin{pmatrix} -1 & 2 \\ -2 & 3 \end{pmatrix} $. Final answer: $ \boxed{\begin{pmatrix} -1 & 2 \\ -2 & 3 \end{pmatrix}} $.
- Question: Find $ y $ such that the vectors $ \begin{pmatrix} 2 \\ y \\ -1 \end{pmatrix} $ and $ \begin{pmatrix} 3 \\ -2 \\ y \end{pmatrix} $ are orthogonal.
- 2(3) + y(-2) + (-1)(y) = 6 - 2y - y = 6 - 3y = 0.
- Solving $ 6 - 3y = 0 $ gives $ y = 2 $. Final answer: $ \boxed{2} $.
- Question: If $ \mathbf{a}, \mathbf{b}, \mathbf{c} $ are unit vectors with $ \mathbf{a} \cdot \mathbf{b} = \frac{1}{2} $, $ \mathbf{b} \cdot \mathbf{c} = \frac{\sqrt{3}}{2} $, find the maximum value of $ \mathbf{a} \cdot \mathbf{c} $.
