Question

Let $\bar{a} \times \bar{b} + \bar{b} = \bar{c} \times \bar{a} + \bar{b} \times \bar{c}$ where $|\bar{a}| = |\bar{b}|$ then identify the correct statement(s)

• A. $\bar{a}.\bar{b} = 0$
• B. $\bar{b}.\bar{c} = 0$
• C. $\bar{c}.\bar{a} = 0$
• D. $\bar{c} = \bar{0}$

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The given equation is $\bar{a} \times \bar{b} + \bar{b} = \bar{c} \times \bar{a} + \bar{b} \times \bar{c}$, with the condition $|\bar{a}| = |\bar{b}|$.

Rearranging the equation: $\bar{a} \times \bar{b} - \bar{c} \times \bar{a} - \bar{b} \times \bar{c} + \bar{b} = \bar{0}$

Using the property $\bar{x} \times \bar{y} = -\bar{y} \times \bar{x}$, we have $\bar{c} \times \bar{a} = -\bar{a} \times \bar{c}$ and $\bar{b} \times \bar{c} = -\bar{c} \times \bar{b}$. Substituting these into the equation: $\bar{a} \times \bar{b} + \bar{a} \times \bar{c} + \bar{c} \times \bar{b} + \bar{b} = \bar{0}$ $\bar{a} \times (\bar{b} + \bar{c}) + \bar{c} \times \bar{b} + \bar{b} = \bar{0}$

Taking the dot product of the original equation with $\bar{b}$: $\bar{b} \cdot (\bar{a} \times \bar{b} + \bar{b}) = \bar{b} \cdot (\bar{c} \times \bar{a} + \bar{b} \times \bar{c})$ $\bar{b} \cdot (\bar{a} \times \bar{b}) + \bar{b} \cdot \bar{b} = \bar{b} \cdot (\bar{c} \times \bar{a}) + \bar{b} \cdot (\bar{b} \times \bar{c})$

Using the property of scalar triple product $\bar{x} \cdot (\bar{y} \times \bar{x}) = 0$ and $\bar{x} \cdot (\bar{x} \times \bar{y}) = 0$: $0 + |\bar{b}|^2 = \bar{b} \cdot (\bar{c} \times \bar{a}) + 0$ $|\bar{b}|^2 = \bar{b} \cdot (\bar{c} \times \bar{a})$

Using the cyclic property of scalar triple product $\bar{b} \cdot (\bar{c} \times \bar{a}) = \bar{a} \cdot (\bar{b} \times \bar{c})$. So, $|\bar{b}|^2 = \bar{a} \cdot (\bar{b} \times \bar{c})$.

Taking the dot product of the original equation with $\bar{a}$: $\bar{a} \cdot (\bar{a} \times \bar{b} + \bar{b}) = \bar{a} \cdot (\bar{c} \times \bar{a} + \bar{b} \times \bar{c})$ $\bar{a} \cdot (\bar{a} \times \bar{b}) + \bar{a} \cdot \bar{b} = \bar{a} \cdot (\bar{c} \times \bar{a}) + \bar{a} \cdot (\bar{b} \times \bar{c})$ $0 + \bar{a} \cdot \bar{b} = 0 + \bar{a} \cdot (\bar{b} \times \bar{c})$ $\bar{a} \cdot \bar{b} = \bar{a} \cdot (\bar{b} \times \bar{c})$

Comparing the results from dot products with $\bar{a}$ and $\bar{b}$: $\bar{a} \cdot \bar{b} = \bar{a} \cdot (\bar{b} \times \bar{c})$ and $|\bar{b}|^2 = \bar{a} \cdot (\bar{b} \times \bar{c})$. Therefore, $\bar{a} \cdot \bar{b} = |\bar{b}|^2$. Since $|\bar{a}| = |\bar{b}|$, we have $\bar{a} \cdot \bar{b} = |\bar{a}|^2$.

From $\bar{a} \cdot \bar{b} = |\bar{b}|^2$, if $\bar{a} \cdot \bar{b} = 0$, then $|\bar{b}|^2 = 0$, which implies $\bar{b} = \bar{0}$. Since $|\bar{a}| = |\bar{b}|$, $\bar{a} = \bar{0}$.

If $\bar{a} = \bar{0}$ and $\bar{b} = \bar{0}$, the original equation becomes $\bar{0} + \bar{0} = \bar{0} + \bar{0}$, which is true. In this case, $\bar{a} \cdot \bar{b} = 0 \cdot 0 = 0$.

Also, taking the dot product of the original equation with $\bar{c}$: $\bar{c} \cdot (\bar{a} \times \bar{b} + \bar{b}) = \bar{c} \cdot (\bar{c} \times \bar{a} + \bar{b} \times \bar{c})$ $\bar{c} \cdot (\bar{a} \times \bar{b}) + \bar{c} \cdot \bar{b} = \bar{c} \cdot (\bar{c} \times \bar{a}) + \bar{c} \cdot (\bar{b} \times \bar{c})$ $\bar{c} \cdot (\bar{a} \times \bar{b}) + \bar{c} \cdot \bar{b} = 0 + 0$ $\bar{c} \cdot (\bar{a} \times \bar{b}) + \bar{c} \cdot \bar{b} = 0$.

If $\bar{a} = \bar{b} = \bar{0}$, then the original equation is satisfied for any $\bar{c}$.

In this case: (A) $\bar{a} \cdot \bar{b} = 0 \cdot 0 = 0$. True. (B) $\bar{b} \cdot \bar{c} = 0 \cdot \bar{c} = 0$. True. (C) $\bar{c} \cdot \bar{a} = \bar{c} \cdot 0 = 0$. True. (D) $\bar{c} = \bar{0}$. Not necessarily true.

Therefore, statements A, B, and C are correct.