If $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})$, where $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ are any three vectors such that $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}} \neq 0$, $\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} \neq 0$, then $\overline{\mathrm{a}}$ and $\overline{\mathrm{c}}$ are

(A) inclined at an angle of $\pi / 3$ between them

(B) inclined at an angle of $\pi / 6$ between them

Vector Algebra — JEE Main · 2006 · 30 Apr 2006 · Offline Question with Solution

(\overline{a} \times \overline{b}) \times \overline{c} = \overline{a} \times (\overline{b} \times \overline{c})

, where

\overline{a}, \overline{b}

and

\overline{c}

are any three vectors such that

\overline{a} \cdot \overline{b} \neq = 0

\overline{b} \cdot \overline{c} \neq = 0

, then

\overline{a}

and

\overline{c}

are

Answer: (D)

Expand each side using the vector triple product rule. The left side is

(\overset{a}{ˉ} \times \overset{ˉ}{b}) \times \overset{c}{ˉ} = (\overset{a}{ˉ} \cdot \overset{c}{ˉ}) \overset{ˉ}{b} - (\overset{ˉ}{b} \cdot \overset{c}{ˉ}) \overset{a}{ˉ}

, while the right side is

\overset{a}{ˉ} \times (\overset{ˉ}{b} \times \overset{c}{ˉ}) = (\overset{a}{ˉ} \cdot \overset{c}{ˉ}) \overset{ˉ}{b} - (\overset{a}{ˉ} \cdot \overset{ˉ}{b}) \overset{c}{ˉ}

. Setting them equal cancels the common term

(\overset{a}{ˉ} \cdot \overset{c}{ˉ}) \overset{ˉ}{b}

and leaves

(\overset{ˉ}{b} \cdot \overset{c}{ˉ}) \overset{a}{ˉ} = (\overset{a}{ˉ} \cdot \overset{ˉ}{b}) \overset{c}{ˉ}

. Because

\overset{a}{ˉ} \cdot \overset{ˉ}{b} \neq = 0

and

\overset{ˉ}{b} \cdot \overset{c}{ˉ} \neq = 0

, this forces

\overset{a}{ˉ}

to be a scalar multiple of

\overset{c}{ˉ}

, so

\overset{a}{ˉ}

and

\overset{c}{ˉ}

are parallel.

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option