Question

Let a and b be two non-zero vectors perpendicular to each other and |a| = |b|. If |a × b| = |a|, then the angle between the vectors (a + b + (a × b)) and a is equal to :

• A. cos⁻¹(1/√3)
• B. sin⁻¹(1/√3)
• C. cos⁻¹(1/√2)
• D. sin⁻¹(1/√6)

Hint:

$a, b,$ and $a \times b$ form an orthogonal basis. Any vector $V$ in this space has a magnitude $|V| = \sqrt{|a|^2 + |b|^2 + |a \times b|^2}$.

Answer 1

The Correct Option is A

Step 1: Let $|a| = |b| = k$. Since $a \perp b$, $|a \times b| = |a||b| \sin 90^\circ = k^2$. Given $|a \times b| = |a| \implies k^2 = k \implies k=1$. So $|a|=|b|=|a \times b|=1$.
Step 2: Let $V = a + b + (a \times b)$. Since $a, b,$ and $(a \times b)$ are mutually perpendicular, $|V|^2 = |a|^2 + |b|^2 + |a \times b|^2 = 1 + 1 + 1 = 3 \implies |V| = \sqrt{3}$.
Step 3: Let $\theta$ be the angle between $V$ and $a$. $\cos \theta = \frac{V \cdot a}{|V||a|} = \frac{(a + b + (a \times b)) \cdot a}{\sqrt{3} \cdot 1}$. $\cos \theta = \frac{a \cdot a + b \cdot a + (a \times b) \cdot a}{\sqrt{3}} = \frac{1 + 0 + 0}{\sqrt{3}} = 1/\sqrt{3}$. $\theta = \cos^{-1}(1/\sqrt{3})$.