Chapter 2: Problem 50
If three vectors \(x \overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}}+3 \vec{c},-2 \overrightarrow{\mathbf{a}}+y \overrightarrow{\mathbf{b}}-4 \overrightarrow{\mathbf{c}}\) and \(-z \vec{b}+2 \vec{c}\) are coplanar, where \(\vec{a} \cdot \vec{b} \cdot \vec{c}\) are unit (or any) vectors, then (a) \(x y+3 z x-3 z=4\) (b) \(2 x y-2 z x-3 z-4=0\) (c) \(4 x y-3 z x+3 z=4\) (d) \(x y-2 z x+3 z-4=0\)
Short Answer
Step by step solution
Understanding Coplanarity Condition
Compute the Cross Product \( \vec{v_2} \times \vec{v_3} \)
Compute Dot Product \( \vec{v_1} \cdot (\vec{v_2} \times \vec{v_3}) \)
Simplify and Solve the Equation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Scalar Triple Product
This can be expressed as follows:
- Calculate the cross product of two vectors, creating a new vector.
- Apply the dot product between the resulting vector and the third vector.
Cross Product
For vectors \( \vec{a} \) and \( \vec{b} \), the cross product \( \vec{a} \times \vec{b} \) creates a vector perpendicular to both, calculated using a 3x3 determinant:
- The determinant consists of unit vectors \( \vec{i}, \vec{j}, \vec{k} \).
- Values of the original vectors are inside the determinant.
Dot Product
Given vectors are unit vectors with
- \( \vec{a} \cdot \vec{a} = 1 \)
- \( \vec{b} \cdot \vec{b} = 1 \)
- \( \vec{c} \cdot \vec{c} = 1 \)
Determinant Expansion
The matrix is arranged as follows:
- Top row with unit vectors \( \vec{i}, \vec{j}, \vec{k} \)
- Subsequent rows represent the components of two vectors.