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Home » Final answer: $ \boxed30 - 12\sqrt6 $ - Dachbleche24

Final answer: $ \boxed30 - 12\sqrt6 $ - Dachbleche24

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 Solution: The matrix $\mathbf{M}$ is constructed by placing the images of the standard basis vectors as its columns. Thus, $\mathbf{M} = \begin{pmatrix} 2 & 3 \\ -1 & 4 \end{pmatrix}$. Verifying, $\mathbf{M} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ and $\mathbf{M} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, confirming correctness. $\boxed{\begin{pmatrix} 2 & 3 \\ -1 & 4 \end{pmatrix}}$

📌 Question: An environmental consultant models a river's flow as the line $y = -\frac{1}{2}x + 5$. Find the point on this line closest to the pollution source at $(4, 3)$.

📌 Solution: The closest point is the projection of $(4, 3)$ onto the line. The formula for the projection of a point $(x_0, y_0)$ onto $ax + by + c = 0$ is used. Rewriting the line as $\frac{1}{2}x + y - 5 = 0$, we compute the projection. Alternatively, parametrize the line and minimize distance. Let $x = t$, then $y = -\frac{1}{2}t + 5$. The squared distance to $(4, 3)$ is $(t - 4)^2 + \left(-\frac{1}{2}t + 5 - 3\right)^2 = (t - 4)^2 + \left(-\frac{1}{2}t + 2\right)^2$. Expanding: $t^2 - 8t + 16 + \frac{1}{4}t^2 - 2t + 4 = \frac{5}{4}t^2 - 10t + 20$. Taking derivative and setting to zero: $\frac{5}{2}t - 10 = 0 \Rightarrow t = 4$. Substituting back, $y = -\frac{1}{2}(4) + 5 = 3$. Thus, the closest point is $(4, 3)$, which lies on the line. $\boxed{(4, 3)}$

📌 Question: A hydrologist models groundwater flow with vectors $\mathbf{a} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$. Find the angle between these flow directions.

📌 Solution: The angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$ is given by $\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}$. Compute the dot product: $2(1) + (-3)(4) = 2 - 12 = -10$. Compute magnitudes: $\|\mathbf{a}\| = \sqrt{2^2 + (-3)^2} = \sqrt{13}$, $\|\mathbf{b}\| = \sqrt{1^2 + 4^2} = \sqrt{17}$. Thus, $\cos\theta = \frac{-10}{\sqrt{13}\sqrt{17}}$. Rationalizing, $\theta = \arccos\left(-\frac{10}{\sqrt{221}}\right)$. $\boxed{\arccos\left(-\dfrac{10}{\sqrt{221}}\right)}$

📌 Question: If $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are unit vectors in hydrology modeling, find the maximum value of $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$.

📌 Solution: The scalar triple product $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$ represents the volume of the parallelepiped formed by the vectors. Its maximum absolute value occurs when the vectors are mutually orthogonal. Since they are unit vectors, the maximum value is $1 \cdot 1 \cdot 1 = 1$. For example, if $\mathbf{u} = \mathbf{v} = \mathbf{w}$ are orthonormal, the

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