Let $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}$ be such that $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=1$. If $a \cos \theta=b \cos \left(\theta+\frac{2 \pi}{3}\right)=c \cos \left(\theta+\frac{4 \pi}{3}\right)$, where $\theta=\frac{\pi}{9}$, then the angle between the vectors $a \hat{i}+b \hat{j}+c \hat{k}$ and $b \hat{i}+c \hat{j}+a \hat{k}$ is:

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Vector Algebra

Let

a, b, c \in R

be such that

a^{2} + b^{2} + c^{2} = 1

. If

a cos θ = b cos (θ + \frac{2 π}{3}) = c cos (θ + \frac{4 π}{3})

, where

θ = \frac{π}{9}

, then the angle between the vectors

a \hat{i} + b \hat{j} + c \hat{k}

and

b \hat{i} + c \hat{j} + a \hat{k}

is:

Answer: (D)

Let the common value be

k

, so

a = \frac{k}{cos θ}

b = \frac{k}{cos ( θ + 2 π /3 )}

and

c = \frac{k}{cos ( θ + 4 π /3 )}

. The angle between

a \hat{i} + b \hat{j} + c \hat{k}

and

b \hat{i} + c \hat{j} + a \hat{k}

turns on their dot product

ab + b c + c a

. Taking out the common factor,

ab + b c + c a = k^{2} \frac{cos θ + cos ( θ + 2 π /3 ) + cos ( θ + 4 π /3 )}{cos θ cos ( θ + 2 π /3 ) cos ( θ + 4 π /3 )}

, and three cosines spaced by

2 π /3

sum to zero, so the dot product is zero. The vectors are perpendicular, and the angle between them is

\frac{π}{2}

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option