Let $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are two non-zero vectors perpendicular to each other and $|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|$. If $|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}|$, then the angle between the vectors $(\vec{a}+\vec{b}+(\vec{a} \times \vec{b}))$ and $\vec{a}$ is equal to :

(C) $\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$

(A) $\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$

(B) $\sin^{-1}\left(\frac{1}{\sqrt{6}}\right)$

(D) $\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Vector Algebra

Let

a

and

b

are two non-zero vectors perpendicular to each other and

∣ a ∣ = ∣ b ∣

. If

∣ a \times b ∣ = ∣ a ∣

, then the angle between the vectors

(a + b + (a \times b))

and

a

is equal to :

Answer: (C)

Since

a

and

b

are perpendicular with

∣ a ∣ = ∣ b ∣

a \times b = ∣ a ∣∣ b ∣ = ∣ a ∣^{2}

, and the condition

a \times b = ∣ a ∣

forces

∣ a ∣ = ∣ b ∣ = 1

, so

a \times b = 1

as well. Write

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

. Because

a \cdot b = 0

and

a \times b

is perpendicular to both

a

and

b

c \cdot a = ∣ a ∣^{2} = 1, ∣ c ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + a \times b^{2} = 3

Hence, if

θ

is the angle between

c

and

a

cos θ = \frac{c \cdot a}{∣ c ∣ ∣ a ∣} = \frac{1}{3}

so the angle is

cos^{- 1} (\frac{1}{3})

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option