Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$. If $\lambda=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ and $\vec{d}=\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}$, then the ordered pair, ( $\lambda, \vec{d}$ ) is equal to:

(C) $\left(-\frac{3}{2}, 3(\vec{a} \times \vec{b})\right)$

(A) $\left(\frac{3}{2}, 3 \vec{b} \times \vec{c}\right)$

(B) $\left(\frac{3}{2}, 3 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}\right)$

(D) $\left(-\frac{3}{2}, 3 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right)$

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Vector Algebra

Let

a, b

and

c

be three unit vectors such that

a + b + c = 0

. If

λ = a \cdot b + b \cdot c + c \cdot a

and

d = a \times b + b \times c + c \times a

, then the ordered pair, (

λ, d

) is equal to:

Answer: (C)

Squaring

a + b + c = 0

and using the unit magnitudes gives

3 + 2 λ = 0

, so

λ = - \frac{3}{2}

. For

d

, substitute

c = - (a + b)

: then

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

and

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

, so

d = a \times b + a \times b + a \times b = 3 (a \times b)

. The ordered pair is

(- \frac{3}{2}, 3 (a \times b))

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option