Math Problem Statement
Solution
To solve the given problem, we need to address two parts:
1. Find a vector of magnitude 14 units orthogonal to vectors and .
An orthogonal vector to both and can be found by calculating the cross product of and .
Given the vectors:
The cross product is calculated as:
\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 3 & 0 \\ 1 & 0 & 2 \end{vmatrix}$$ Expanding this determinant: $$\mathbf{a} \times \mathbf{b} = \mathbf{i} \left( 3 \times 2 - 0 \times 0 \right) - \mathbf{j} \left( -1 \times 2 - 0 \times 1 \right) + \mathbf{k} \left( -1 \times 0 - 3 \times 1 \right)$$ $$\mathbf{a} \times \mathbf{b} = \mathbf{i}(6) - \mathbf{j}(-2) + \mathbf{k}(-3)$$ $$\mathbf{a} \times \mathbf{b} = 6\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}$$ Thus, the vector orthogonal to both $$\mathbf{a}$$ and $$\mathbf{b}$$ is $$\mathbf{v} = 6\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}$$. Next, we scale this vector to have a magnitude of 14 units. The magnitude of $$\mathbf{v}$$ is: $$|\mathbf{v}| = \sqrt{6^2 + 2^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$$ To get a vector with magnitude 14, we multiply $$\mathbf{v}$$ by 2 (since $$\frac{14}{7} = 2$$): $$\mathbf{v}_{14} = 2(6\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) = 12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}$$ Thus, the vector of magnitude 14 units orthogonal to both $$\mathbf{a}$$ and $$\mathbf{b}$$ is: $$\boxed{12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}}$$ ### 2. Find the angle between the vectors $$\mathbf{a}$$ and $$\mathbf{b}$$. The formula to find the angle $$\theta$$ between two vectors is: $$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}$$ First, compute the dot product $$\mathbf{a} \cdot \mathbf{b}$$: $$\mathbf{a} \cdot \mathbf{b} = (-1)(1) + (3)(0) + (0)(2) = -1$$ Now, find the magnitudes of $$\mathbf{a}$$ and $$\mathbf{b}$$: $$|\mathbf{a}| = \sqrt{(-1)^2 + 3^2 + 0^2} = \sqrt{1 + 9} = \sqrt{10}$$ $$|\mathbf{b}| = \sqrt{1^2 + 0^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$$ Substitute into the cosine formula: $$\cos \theta = \frac{-1}{\sqrt{10} \times \sqrt{5}} = \frac{-1}{\sqrt{50}} = \frac{-1}{5\sqrt{2}}$$ Thus, the angle $$\theta$$ is: $$\theta = \cos^{-1} \left( \frac{-1}{5\sqrt{2}} \right)$$ You can further simplify or approximate using a calculator, but this is the exact form for the angle. Would you like to know more details about any step, or would you like me to further simplify this? ### Follow-up Questions: 1. How do we verify the orthogonality of the resulting vector to the original ones? 2. What is the geometric interpretation of the cross product in this context? 3. Can you explain why the cross product results in a vector perpendicular to both input vectors? 4. How do you find the angle between vectors without a calculator? 5. How does changing the magnitude of a vector affect its direction? ### Tip: Remember, the cross product of two vectors is always perpendicular to both, but the direction depends on the right-hand rule.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Vector Algebra
Cross Product
Dot Product
Magnitude of Vectors
Angles Between Vectors
Formulas
Cross Product: a × b = |a| |b| sin(θ) n̂
Dot Product: a · b = |a| |b| cos(θ)
Magnitude of a Vector: |v| = √(x² + y² + z²)
Angle Between Vectors: cos(θ) = (a · b) / (|a| |b|)
Theorems
Orthogonality Theorem: Two vectors are orthogonal if their dot product is zero.
Cross Product Results in a Vector Perpendicular to Two Input Vectors
Suitable Grade Level
Grades 10-12
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