Question

$\overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c}, \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{d}$, where $\overrightarrow{c} \neq 0$, then

• A. $|\overrightarrow{a}| = |\overrightarrow{c}|, |\overrightarrow{b}| = 1$
• B. $|\overrightarrow{a}| = |\overrightarrow{b}|, |\overrightarrow{c}| = 1$
• C. $|\overrightarrow{b}| = |\overrightarrow{c}|, |\overrightarrow{a}| = 1$
• D. $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| = 1$

Answer 1

Answer: (A)

(1)

Answer 2

Given the vector equations:

1. $\overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c}$
2. $\overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{d}$

Also, $\overrightarrow{c} \neq 0$.

From equation (1), $\overrightarrow{c}$ is perpendicular to $\overrightarrow{b}$. Therefore, the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $90^\circ$. From equation (2), the magnitude of $\overrightarrow{d}$ is given by: $|\overrightarrow{d}| = |\overrightarrow{b}| |\overrightarrow{c}| \sin(90^\circ) = |\overrightarrow{b}| |\overrightarrow{c}|$ (Equation A)

Substitute $\overrightarrow{c} = \overrightarrow{a} \times \overrightarrow{b}$ into the second equation: $\overrightarrow{d} = \overrightarrow{b} \times (\overrightarrow{a} \times \overrightarrow{b})$ Using the vector triple product identity $\overrightarrow{A} \times (\overrightarrow{B} \times \overrightarrow{C}) = (\overrightarrow{A} \cdot \overrightarrow{C})\overrightarrow{B} - (\overrightarrow{A} \cdot \overrightarrow{B})\overrightarrow{C}$: $\overrightarrow{d} = (\overrightarrow{b} \cdot \overrightarrow{b})\overrightarrow{a} - (\overrightarrow{b} \cdot \overrightarrow{a})\overrightarrow{b}$ $\overrightarrow{d} = |\overrightarrow{b}|^2 \overrightarrow{a} - (\overrightarrow{a} \cdot \overrightarrow{b})\overrightarrow{b}$

For the options to be uniquely determined, we need an additional constraint. A common implicit constraint in such problems is that $\overrightarrow{d}$ is parallel to $\overrightarrow{a}$. If $\overrightarrow{d}$ is parallel to $\overrightarrow{a}$, then the component of $\overrightarrow{d}$ along $\overrightarrow{b}$ must be zero. This means $(\overrightarrow{a} \cdot \overrightarrow{b})\overrightarrow{b} = \overrightarrow{0}$. Since $\overrightarrow{b} \neq \overrightarrow{0}$ (otherwise $\overrightarrow{c} = \overrightarrow{0}$), it implies $\overrightarrow{a} \cdot \overrightarrow{b} = 0$. This means $\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$ (i.e., the angle between them, $\theta_{ab}$, is $90^\circ$).

Under the condition $\overrightarrow{a} \perp \overrightarrow{b}$:

1. The magnitude of $\overrightarrow{c}$ becomes: $|\overrightarrow{c}| = |\overrightarrow{a}| |\overrightarrow{b}| \sin \theta_{ab} = |\overrightarrow{a}| |\overrightarrow{b}| \sin(90^\circ) = |\overrightarrow{a}| |\overrightarrow{b}|$ (Equation B)
2. The expression for $\overrightarrow{d}$ simplifies to: $\overrightarrow{d} = |\overrightarrow{b}|^2 \overrightarrow{a} - (0)\overrightarrow{b} = |\overrightarrow{b}|^2 \overrightarrow{a}$ Taking the magnitude, $|\overrightarrow{d}| = |\overrightarrow{b}|^2 |\overrightarrow{a}|$ (Equation C)

Now, equate Equation A and Equation C: $|\overrightarrow{b}| |\overrightarrow{c}| = |\overrightarrow{b}|^2 |\overrightarrow{a}|$ Since $|\overrightarrow{b}| \neq 0$, we can divide by $|\overrightarrow{b}|$: $|\overrightarrow{c}| = |\overrightarrow{b}| |\overrightarrow{a}|$ This relation is identical to Equation B, confirming consistency.

Now, let's check which option satisfies these conditions. Consider option (1): $|\overrightarrow{a}| = |\overrightarrow{c}|$ and $|\overrightarrow{b}| = 1$. Substitute these into the derived relation $|\overrightarrow{c}| = |\overrightarrow{a}| |\overrightarrow{b}|$: $|\overrightarrow{a}| = |\overrightarrow{a}| (1)$ $|\overrightarrow{a}| = |\overrightarrow{a}|$ This is consistent. Therefore, option (1) is a possible solution under the implicit assumption that $\overrightarrow{d}$ is parallel to $\overrightarrow{a}$.