Answer on Question #56115 – Math - Vector Calculus
Question
Simplify
( A + B ) . ( B + C ) × ( C + A ) (A + B). (B + C) \times (C + A) ( A + B ) . ( B + C ) × ( C + A )
a). B × C B \times C B × C
b). 2. A × B 2. A \times B 2. A × B
c). 2. A × C 2. A \times C 2. A × C
d). A × B × C A \times B \times C A × B × C
Solution
By the properties of cross product, calculate
( A ⃗ + B ⃗ ) ⋅ ( ( B ⃗ + C ⃗ ) × ( C ⃗ + A ⃗ ) ) = ( A ⃗ + B ⃗ ) ⋅ ( B ⃗ × C ⃗ + B ⃗ × A ⃗ + C ⃗ × C ⃗ + C ⃗ × A ⃗ ) = ( A ⃗ + B ⃗ ) ⋅ ( B ⃗ × C ⃗ + B ⃗ × A ⃗ + 0 + C ⃗ × A ⃗ ) = A ⃗ ⋅ B ⃗ × C ⃗ + A ⃗ ⋅ B ⃗ × A ⃗ + A ⃗ ⋅ C ⃗ × A ⃗ + B ⃗ ⋅ B ⃗ × C ⃗ + B ⃗ ⋅ B ⃗ × A ⃗ + B ⃗ ⋅ C ⃗ × A ⃗ = A ⃗ ⋅ B ⃗ × C ⃗ + 0 + 0 + 0 + 0 + B ⃗ ⋅ C ⃗ × A ⃗ = 2 ⋅ A ⃗ ⋅ B ⃗ × C ⃗ . \begin{array}{l}
(\vec{A} + \vec{B}) \cdot \left((\vec{B} + \vec{C}) \times (\vec{C} + \vec{A})\right) = (\vec{A} + \vec{B}) \cdot (\vec{B} \times \vec{C} + \vec{B} \times \vec{A} + \vec{C} \times \vec{C} + \vec{C} \times \vec{A}) \\
= (\vec{A} + \vec{B}) \cdot (\vec{B} \times \vec{C} + \vec{B} \times \vec{A} + 0 + \vec{C} \times \vec{A}) \\
= \vec{A} \cdot \vec{B} \times \vec{C} + \vec{A} \cdot \vec{B} \times \vec{A} + \vec{A} \cdot \vec{C} \times \vec{A} + \vec{B} \cdot \vec{B} \times \vec{C} + \vec{B} \cdot \vec{B} \times \vec{A} + \vec{B} \cdot \vec{C} \times \vec{A} \\
= \vec{A} \cdot \vec{B} \times \vec{C} + 0 + 0 + 0 + 0 + \vec{B} \cdot \vec{C} \times \vec{A} = 2 \cdot \vec{A} \cdot \vec{B} \times \vec{C}.
\end{array} ( A + B ) ⋅ ( ( B + C ) × ( C + A ) ) = ( A + B ) ⋅ ( B × C + B × A + C × C + C × A ) = ( A + B ) ⋅ ( B × C + B × A + 0 + C × A ) = A ⋅ B × C + A ⋅ B × A + A ⋅ C × A + B ⋅ B × C + B ⋅ B × A + B ⋅ C × A = A ⋅ B × C + 0 + 0 + 0 + 0 + B ⋅ C × A = 2 ⋅ A ⋅ B × C .
Answer. 2 ⋅ A ⃗ ⋅ B ⃗ × C ⃗ 2 \cdot \vec{A} \cdot \vec{B} \times \vec{C} 2 ⋅ A ⋅ B × C
www.AssignmentExpert.com
Comments