Math Problem Statement
A unit vector which is perpendicular to i+2j-2k and -i +2j+ 2k is
Solution
To find a unit vector that is perpendicular to both and , we use the cross product. The result of the cross product of two vectors is a vector perpendicular to both. After calculating the cross product, we normalize the resulting vector to make it a unit vector.
Step 1: Compute the cross product
The cross product formula is:
\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -2 \\ -1 & 2 & 2 \end{vmatrix}$$ Expanding the determinant: $$\mathbf{a} \times \mathbf{b} = \mathbf{i} \begin{vmatrix} 2 & -2 \\ 2 & 2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -2 \\ -1 & 2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ -1 & 2 \end{vmatrix}$$ ### Step 2: Calculate each minor determinant 1. $$\begin{vmatrix} 2 & -2 \\ 2 & 2 \end{vmatrix} = (2)(2) - (2)(-2) = 4 + 4 = 8$$ 2. $$\begin{vmatrix} 1 & -2 \\ -1 & 2 \end{vmatrix} = (1)(2) - (-1)(-2) = 2 - 2 = 0$$ 3. $$\begin{vmatrix} 1 & 2 \\ -1 & 2 \end{vmatrix} = (1)(2) - (-1)(2) = 2 + 2 = 4$$ Substitute these into the formula: $$\mathbf{a} \times \mathbf{b} = 8\mathbf{i} - 0\mathbf{j} + 4\mathbf{k} = 8\mathbf{i} + 4\mathbf{k}$$ ### Step 3: Normalize the result The magnitude of $$\mathbf{a} \times \mathbf{b}$$ is: $$\|\mathbf{a} \times \mathbf{b}\| = \sqrt{8^2 + 0^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$$ The unit vector is: $$\mathbf{u} = \frac{\mathbf{a} \times \mathbf{b}}{\|\mathbf{a} \times \mathbf{b}\|} = \frac{1}{4\sqrt{5}} (8\mathbf{i} + 4\mathbf{k}) = \frac{2}{\sqrt{5}}\mathbf{i} + \frac{1}{\sqrt{5}}\mathbf{k}$$ ### Final Answer: The unit vector is: $$\mathbf{u} = \frac{2}{\sqrt{5}}\mathbf{i} + \frac{1}{\sqrt{5}}\mathbf{k}$$ --- Do you want a step-by-step explanation of any part? Let me know! --- ### Related Questions: 1. How to compute the cross product of two vectors geometrically? 2. What is the significance of a unit vector in physics and mathematics? 3. How to verify if two vectors are perpendicular or parallel? 4. What are alternative methods to find a vector perpendicular to two given vectors? 5. What is the relationship between the magnitude of the cross product and the sine of the angle between two vectors? ### Tip: The magnitude of the cross product gives the area of the parallelogram formed by two vectors.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Vector Algebra
Cross Product
Unit Vector
Formulas
Cross product: a × b = |i j k|
|a1 a2 a3|
|b1 b2 b3|
Unit vector: u = v / |v|
Magnitude of a vector: |v| = √(v1^2 + v2^2 + v3^2)
Theorems
The cross product of two vectors is perpendicular to both.
A unit vector has a magnitude of 1.
Suitable Grade Level
Grades 11-12
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