Vectors a = 2i + 3j - 4k and b = 4i + 2j + 3k represent the two adjacent sides of a triangle. The magnitude of the area of the triangle and the unit vector perpendicular to both a... Vectors a = 2i + 3j - 4k and b = 4i + 2j + 3k represent the two adjacent sides of a triangle. The magnitude of the area of the triangle and the unit vector perpendicular to both a and b respectively, are...
Understand the Problem
The question involves calculating the magnitude of the area of a triangle formed by two vectors and determining the unit vector perpendicular to those vectors.
Answer
$28.93$ and $0.58\mathbf{i} - 0.76\mathbf{j} - 0.27\mathbf{k}$
Answer for screen readers
The magnitude of the area of the triangle is approximately $28.93$, and the unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ is approximately $0.58\mathbf{i} - 0.76\mathbf{j} - 0.27\mathbf{k}$.
Steps to Solve
- Determine the Cross Product of the Vectors
To find the vector perpendicular to both vectors ( \mathbf{a} ) and ( \mathbf{b} ), we calculate the cross product:
$$ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 3 & -4 \ 4 & 2 & 3 \end{vmatrix} $$
Calculating this determinant:
$$ \mathbf{a} \times \mathbf{b} = \mathbf{i} \begin{vmatrix} 3 & -4 \ 2 & 3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & -4 \ 4 & 3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 3 \ 4 & 2 \end{vmatrix} $$
Which gives:
$$ \mathbf{a} \times \mathbf{b} = \mathbf{i}(3 \cdot 3 - (-4) \cdot 2) - \mathbf{j}(2 \cdot 3 - (-4) \cdot 4) + \mathbf{k}(2 \cdot 2 - 3 \cdot 4) $$
- Calculate the Components of the Cross Product
Calculating each component:
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For ( \mathbf{i} ): $$ 9 + 8 = 17 $$
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For ( \mathbf{j} ): $$ 6 + 16 = 22 \quad \text{(note the sign change from -22)} $$
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For ( \mathbf{k} ): $$ 4 - 12 = -8 $$
Thus, we have:
$$ \mathbf{a} \times \mathbf{b} = 17\mathbf{i} - 22\mathbf{j} - 8\mathbf{k} $$
- Calculate Magnitude of the Area of the Triangle
The area ( A ) of the triangle formed by the vectors can be calculated as:
$$ A = \frac{1}{2} |\mathbf{a} \times \mathbf{b}| $$
First, compute the magnitude:
$$ |\mathbf{a} \times \mathbf{b}| = \sqrt{17^2 + (-22)^2 + (-8)^2} $$
- Calculate Each Squared Component
Calculating the squares:
$$ 17^2 = 289, \quad (-22)^2 = 484, \quad (-8)^2 = 64 $$
Combine these:
$$ |\mathbf{a} \times \mathbf{b}| = \sqrt{289 + 484 + 64} = \sqrt{837} $$
- Final Area Calculation
Now find the area:
$$ A = \frac{1}{2} \sqrt{837} \approx 28.93 $$
- Find the Unit Vector
Finally, the unit vector ( \mathbf{u} ) in the direction of ( \mathbf{a} \times \mathbf{b} ) is given by:
$$ \mathbf{u} = \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|} $$
Calculating the components:
$$ \mathbf{u} = \frac{17\mathbf{i} - 22\mathbf{j} - 8\mathbf{k}}{\sqrt{837}} $$
The magnitude of the area of the triangle is approximately $28.93$, and the unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ is approximately $0.58\mathbf{i} - 0.76\mathbf{j} - 0.27\mathbf{k}$.
More Information
The area of a triangle formed by two vectors is half the magnitude of the cross product of those vectors. The unit vector perpendicular to both vectors indicates direction and allows for understanding the orientation of that plane in three dimensions.
Tips
- Miscalculating the Determinant: Ensure each component of the cross product is calculated correctly, paying close attention to sign changes.
- Forgetting the Area Factor: Remember that the area is half the magnitude of the cross product, which is often overlooked.
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