Question

Let \( \vec{\alpha}, \vec{\beta}, \vec{\gamma} \) be three non-zero vectors, such that \( \vec{\alpha} \times \vec{\beta} = \vec{\gamma} \) and \( \vec{\beta} \times \vec{\gamma} = \vec{\alpha} \). Show that \( \vec{\alpha}, \vec{\beta}, \vec{\gamma} \) are mutually perpendicular vectors such that \( |\vec{\beta}| = 1 \) and \( |\vec{\alpha}| = |\vec{\gamma}| \).

Find projection of vector \( \vec{\alpha} \) on \( \vec{\beta} \). Give reasons to support your answer.

Hint:

Each cross product is perpendicular to both its factors, and the magnitudes give \( |\vec{\beta}|^2 = 1 \). The projection is 0.

Answer 1

Step 1: Use the meaning of a cross product:
The vector \( \vec{a} \times \vec{b} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \).
From \( \vec{\alpha} \times \vec{\beta} = \vec{\gamma} \), the vector \( \vec{\gamma} \) is perpendicular to \( \vec{\alpha} \) and to \( \vec{\beta} \).
From \( \vec{\beta} \times \vec{\gamma} = \vec{\alpha} \), the vector \( \vec{\alpha} \) is perpendicular to \( \vec{\beta} \) and to \( \vec{\gamma} \).
All vectors are non-zero, so the three pairs \( (\vec{\alpha}, \vec{\beta}) \), \( (\vec{\beta}, \vec{\gamma}) \), \( (\vec{\gamma}, \vec{\alpha}) \) are each perpendicular.
Hence \( \vec{\alpha}, \vec{\beta}, \vec{\gamma} \) are mutually perpendicular.

Step 2: Compare the magnitudes:
Since the vectors in each cross product are perpendicular, \( \sin 90^\circ = 1 \).
From the first relation, \( |\vec{\gamma}| = |\vec{\alpha}||\vec{\beta}| \).
From the second relation, \( |\vec{\alpha}| = |\vec{\beta}||\vec{\gamma}| \).

Step 3: Find the size of beta:
Put the first result into the second.
\[ |\vec{\alpha}| = |\vec{\beta}|\,|\vec{\alpha}||\vec{\beta}| = |\vec{\beta}|^2 |\vec{\alpha}| \]
Because \( |\vec{\alpha}| \neq 0 \), we get \( |\vec{\beta}|^2 = 1 \), so \( |\vec{\beta}| = 1 \).
Then \( |\vec{\gamma}| = |\vec{\alpha}| \cdot 1 = |\vec{\alpha}| \).
This proves both required facts.

Step 4: Find the projection:
The projection of \( \vec{\alpha} \) on \( \vec{\beta} \) is \( \dfrac{\vec{\alpha}\cdot\vec{\beta}}{|\vec{\beta}|} \).
The vectors are perpendicular, so \( \vec{\alpha}\cdot\vec{\beta} = 0 \).
\[ \text{Projection} = \frac{0}{1} = 0 \]

Final Answer:
The projection of \( \vec{\alpha} \) on \( \vec{\beta} \) is 0, because \( \vec{\alpha} \perp \vec{\beta} \) and so their dot product is zero. \[ \boxed{0} \]