Let $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be two vectors. If $\vec{c}$ is a vector such that $\vec{b} \times \vec{c}=\vec{b} \times \vec{a}$ and $\vec{c} \cdot \vec{a}=0$, then $\vec{c} \cdot \vec{b}$ is equal to:

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Vector Algebra

Let

a = \hat{i} - 2 \hat{j} + \hat{k}

and

b = \hat{i} - \hat{j} + \hat{k}

be two vectors. If

c

is a vector such that

b \times c = b \times a

and

c \cdot a = 0

, then

c \cdot b

is equal to:

Answer: (A)

From

b \times c = b \times a

we get

b \times (c - a) = 0

, so

c - a

is parallel to

b

and

c = a + λ b

. Imposing

c \cdot a = 0

gives

∣ a ∣^{2} + λ (a \cdot b) = 0

; with

∣ a ∣^{2} = 1 + 4 + 1 = 6

and

a \cdot b = 1 + 2 + 1 = 4

, this is

6 + 4 λ = 0

, so

λ = - \frac{3}{2}

. Then

c \cdot b = a \cdot b + λ ∣ b ∣^{2} = 4 - \frac{3}{2} (3) = 4 - \frac{9}{2} = - \frac{1}{2}

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option