Question

For non-zero vectors $\bar{a}, \bar{b}, \bar{c}$ $|(\bar{a} \times \bar{b}).\bar{c}| = |\bar{a}| |\bar{b}| |\bar{c}|$ holds iff.:

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

Answer 1

Answer: (D)

$\bar{a}.\bar{b} = \bar{b}.\bar{c} = \bar{c}.\bar{a} = 0$

Answer 2

The given condition is $|(\bar{a} \times \bar{b}).\bar{c}| = |\bar{a}| |\bar{b}| |\bar{c}|$.

Let $\theta$ be the angle between vectors $\bar{a}$ and $\bar{b}$ ($0 \le \theta \le \pi$). The magnitude of the cross product is $|\bar{a} \times \bar{b}| = |\bar{a}| |\bar{b}| \sin \theta$.

Let $\bar{n} = \bar{a} \times \bar{b}$. The vector $\bar{n}$ is perpendicular to both $\bar{a}$ and $\bar{b}$. Let $\phi$ be the angle between $\bar{n}$ and $\bar{c}$ ($0 \le \phi \le \pi$). The scalar triple product is $(\bar{a} \times \bar{b}).\bar{c} = \bar{n}.\bar{c} = |\bar{n}| |\bar{c}| \cos \phi$.

Substituting these into the given condition: $||\bar{a}| |\bar{b}| \sin \theta |\bar{c}| \cos \phi| = |\bar{a}| |\bar{b}| |\bar{c}|$ $|\bar{a}| |\bar{b}| |\sin \theta| |\bar{c}| |\cos \phi| = |\bar{a}| |\bar{b}| |\bar{c}|$

Since $\bar{a}, \bar{b}, \bar{c}$ are non-zero vectors, their magnitudes $|\bar{a}|, |\bar{b}|, |\bar{c}|$ are positive. Also, since $0 \le \theta \le \pi$, $\sin \theta \ge 0$, so $|\sin \theta| = \sin \theta$. Thus, the equation becomes: $|\bar{a}| |\bar{b}| \sin \theta |\bar{c}| |\cos \phi| = |\bar{a}| |\bar{b}| |\bar{c}|$

We can divide both sides by $|\bar{a}| |\bar{b}| |\bar{c}|$ (since they are non-zero): $\sin \theta |\cos \phi| = 1$

We know that:

1. $0 \le \sin \theta \le 1$ (for $0 \le \theta \le \pi$)
2. $0 \le |\cos \phi| \le 1$ (for $0 \le \phi \le \pi$)

For the product of two numbers, both less than or equal to 1, to be equal to 1, both numbers must be equal to 1. Therefore, we must have:

1. $\sin \theta = 1$
2. $|\cos \phi| = 1$

Let's analyze these two conditions:

1. $\sin \theta = 1$: This implies $\theta = \frac{\pi}{2}$ (or $90^\circ$). Since $\theta$ is the angle between $\bar{a}$ and $\bar{b}$, this means $\bar{a}$ is perpendicular to $\bar{b}$. So, $\bar{a}.\bar{b} = 0$.

2. $|\cos \phi| = 1$: This implies $\cos \phi = 1$ or $\cos \phi = -1$. This means $\phi = 0$ or $\phi = \pi$. Since $\phi$ is the angle between $\bar{n} = \bar{a} \times \bar{b}$ and $\bar{c}$, this means $\bar{c}$ is parallel to $\bar{n}$ (or anti-parallel). As $\bar{n} = \bar{a} \times \bar{b}$ is a vector perpendicular to both $\bar{a}$ and $\bar{b}$, if $\bar{c}$ is parallel to $\bar{n}$, then $\bar{c}$ must also be perpendicular to both $\bar{a}$ and $\bar{b}$. So, $\bar{c} \perp \bar{a} \implies \bar{c}.\bar{a} = 0$. And $\bar{c} \perp \bar{b} \implies \bar{c}.\bar{b} = 0$.

Combining all conditions, for $|(\bar{a} \times \bar{b}).\bar{c}| = |\bar{a}| |\bar{b}| |\bar{c}|$ to hold, we must have: $\bar{a}.\bar{b} = 0$ $\bar{b}.\bar{c} = 0$ $\bar{c}.\bar{a} = 0$

This means that the three vectors $\bar{a}, \bar{b}, \bar{c}$ must be mutually orthogonal (perpendicular to each other).