THE CROSS PRODUCT Find two unit vectors orthogonal
The Correct Answer and Explanation is :
We need to find two unit vectors that are orthogonal (perpendicular) to both given vectors ( \mathbf{a} ) and ( \mathbf{b} ) using the cross product.
Step 1: Compute the Cross Product ( \mathbf{a} \times \mathbf{b} )
Given vectors:
[
\mathbf{a} = (1, 2, 3), \quad \mathbf{b} = (4, 5, 6)
]
The cross product formula for two 3D vectors is:
[
\mathbf{a} \times \mathbf{b} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \
1 & 2 & 3 \
4 & 5 & 6
\end{vmatrix}
]
Expanding along the first row:
[
\mathbf{a} \times \mathbf{b} =
\mathbf{i} \begin{vmatrix} 2 & 3 \ 5 & 6 \end{vmatrix}
- \mathbf{j} \begin{vmatrix} 1 & 3 \ 4 & 6 \end{vmatrix}
- \mathbf{k} \begin{vmatrix} 1 & 2 \ 4 & 5 \end{vmatrix}
]
Computing the 2×2 determinants:
[
\begin{vmatrix} 2 & 3 \ 5 & 6 \end{vmatrix} = (2)(6) – (3)(5) = 12 – 15 = -3
]
[
\begin{vmatrix} 1 & 3 \ 4 & 6 \end{vmatrix} = (1)(6) – (3)(4) = 6 – 12 = -6
]
[
\begin{vmatrix} 1 & 2 \ 4 & 5 \end{vmatrix} = (1)(5) – (2)(4) = 5 – 8 = -3
]
Thus:
[
\mathbf{a} \times \mathbf{b} = (-3, 6, -3)
]
Step 2: Compute the Unit Vectors
The magnitude of ( \mathbf{a} \times \mathbf{b} ) is:
[
|\mathbf{a} \times \mathbf{b}| = \sqrt{(-3)^2 + 6^2 + (-3)^2}
]
[
= \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6}
]
The unit vectors are:
[
\mathbf{u}_1 = \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|}
= \frac{(-3, 6, -3)}{3\sqrt{6}} = \left(\frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}\right)
]
[
\mathbf{u}_2 = -\mathbf{u}_1 = \left(\frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)
]
Conclusion
The two unit vectors orthogonal to both given vectors are:
[
\left(\frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}\right) \quad \text{and} \quad \left(\frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)
]
Explanation (300 Words)
The cross product of two vectors in three-dimensional space results in a vector that is orthogonal to both given vectors. This property is particularly useful in physics, engineering, and computer graphics.
To determine the unit vectors perpendicular to ( \mathbf{a} ) and ( \mathbf{b} ), we first compute their cross product. The determinant method ensures that we correctly find the perpendicular direction. After computing the determinant of the 3×3 matrix, we obtain ( (-3,6,-3) ), which represents a perpendicular vector to both ( \mathbf{a} ) and ( \mathbf{b} ).
Since unit vectors must have a magnitude of 1, we normalize the result by dividing each component by the vector’s magnitude. This process ensures that the resulting vector maintains direction while having a unit length. The magnitude of our cross-product vector is ( 3\sqrt{6} ), leading to the final unit vectors ( \left(\frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}\right) ) and its negative counterpart.
These unit vectors are crucial in many applications, such as computing normal vectors in 3D rendering or defining planes in geometry. By using this method, we can always find two perpendicular unit vectors for any non-parallel 3D vectors.