Let $\vec{a}, \vec{b}$, and $\vec{c}$ be three unit vectors, out of which vectors $\vec{b}$ and $\vec{c}$ are non-parallel. If $\alpha$ and $\beta$ are the angles which vector $\vec{a}$ makes with vectors $\vec{b}$ and $\vec{c}$ respectively and $\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\frac{1}{2} \overrightarrow{\mathrm{~b}}$, then $|\alpha-\beta|$ is equal to :

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Vector Algebra

Let

a, b

, and

c

be three unit vectors, out of which vectors

b

and

c

are non-parallel. If

α

and

β

are the angles which vector

a

makes with vectors

b

and

c

respectively and

a \times (b \times c) = \frac{1}{2} b

, then

∣ α - β ∣

is equal to :

Answer: (A)

Expand

a \times (b \times c) = (a \cdot c) b - (a \cdot b) c

and set it equal to

\frac{1}{2} b

. Since

b

and

c

are non-parallel, matching coefficients forces

a \cdot c = \frac{1}{2}

and

a \cdot b = 0

. With unit vectors

cos β = \frac{1}{2}

gives

β = 6 0^{\circ}

and

cos α = 0

gives

α = 9 0^{\circ}

, so

∣ α - β ∣ = 3 0^{\circ}

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option