Question

$\vec{a}, \vec{b}, \vec{c}, \vec{d}$ be non-zero vectors such that $\vec{a}\times\vec{b}=\vec{c}\times\vec{d}$ and $\vec{a}\times\vec{c}=\vec{b}\times\vec{d}$. Then:

• A. $\vec{a}-\vec{d}$ is parallel to $\vec{b}-\vec{c}$
• B. $\vec{a}-\vec{b}$ is parallel to $\vec{b}-\vec{c}$
• C. $\vec{b}-\vec{c}$ is parallel to $\vec{b}+\vec{c}$
• D. $\vec{a}-\vec{c}$ is parallel to $\vec{b}-\vec{c}$
• E. $\vec{a}+\vec{c}$ is parallel to $\vec{b}+\vec{d}$

Hint:

The cross product $\vec{u} \times \vec{v} = 0$ implies vectors $\vec{u}$ and $\vec{v}$ are collinear or parallel.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given two conditions involving cross products of four vectors. We need to determine the relationship between certain combinations of these vectors. Two vectors are parallel if their cross product is the zero vector. Our goal is to manipulate the given equations to show that the cross product of the vectors in one of the options is zero.
Step 2: Key Formula or Approach:
1. Property of cross product: \( \vec{u} \times \vec{v} = -(\vec{v} \times \vec{u}) \) (Anti-commutative).
2. Distributive property: \( \vec{u} \times (\vec{v} + \vec{w}) = (\vec{u} \times \vec{v}) + (\vec{u} \times \vec{w}) \).
3. Condition for parallelism: Two non-zero vectors \( \vec{P} \) and \( \vec{Q} \) are parallel if and only if \( \vec{P} \times \vec{Q} = \vec{0} \).
Step 3: Detailed Explanation:
We are given:
(1) \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \)
(2) \( \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \)
Let's manipulate the second equation. Using the anti-commutative property, we can write \( \vec{b} \times \vec{d} = -(\vec{d} \times \vec{b}) \). Let's rearrange equation (2):
\[ \vec{a} \times \vec{c} - \vec{b} \times \vec{d} = \vec{0} \] This doesn't seem to lead anywhere directly. Let's try subtracting the two given equations.
Subtract equation (2) from equation (1):
\[ (\vec{a} \times \vec{b}) - (\vec{a} \times \vec{c}) = (\vec{c} \times \vec{d}) - (\vec{b} \times \vec{d}) \] Use the distributive property of the cross product:
\[ \vec{a} \times (\vec{b} - \vec{c}) = (\vec{c} - \vec{b}) \times \vec{d} \] Use the anti-commutative property on the right side: \( (\vec{c} - \vec{b}) = -(\vec{b} - \vec{c}) \).
\[ \vec{a} \times (\vec{b} - \vec{c}) = -(\vec{b} - \vec{c}) \times \vec{d} \] Rearrange the terms to one side:
\[ \vec{a} \times (\vec{b} - \vec{c}) + (\vec{b} - \vec{c}) \times \vec{d} = \vec{0} \] Again, using the anti-commutative property \( (\vec{b} - \vec{c}) \times \vec{d} = -(\vec{d} \times (\vec{b} - \vec{c})) \):
\[ \vec{a} \times (\vec{b} - \vec{c}) - \vec{d} \times (\vec{b} - \vec{c}) = \vec{0} \] Now, use the distributive property in reverse:
\[ (\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = \vec{0} \] This result shows that the cross product of the vector \( (\vec{a} - \vec{d}) \) and the vector \( (\vec{b} - \vec{c}) \) is the zero vector.
Step 4: Final Answer:
Since the cross product of \( (\vec{a} - \vec{d}) \) and \( (\vec{b} - \vec{c}) \) is zero, and we are given that the original vectors are non-zero (implying these combinations are also likely non-zero), the two vectors must be parallel.