Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors mutually perpendicular to each other and have same magnitude. If a vector $\vec{r}$ satisfies $\vec{a} \times\{(\vec{r}-\vec{b}) \times \vec{a}\}+\vec{b} \times\{(\vec{r}-\vec{b}) \times \vec{b}\}+\vec{c} \times\{(\vec{r}-\vec{a}) \times \vec{c}\}=\overrightarrow{0}$, then $\vec{r}$ is equal to

(A) $\frac{1}{2}(\vec{a}+\vec{b}+\vec{c})$

(B) $\frac{1}{3}(\vec{a}+\vec{b}+\vec{c})$

(C) $\frac{1}{3}(2 \vec{a}+\vec{b}-\vec{c})$

(D) $\frac{1}{2}(\vec{a}+\vec{b}+2 \vec{c})$

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Vector Algebra

Let

a, b, c

be three vectors mutually perpendicular to each other and have same magnitude. If a vector

r

satisfies

a \times {(r - b) \times a} + b \times {(r - b) \times b} + c \times {(r - a) \times c} = 0

, then

r

is equal to

Answer: (A)

Using the vector triple product identity

u \times (v \times w) = (u \cdot w) v - (u \cdot v) w

, and noting that

a, b, c

are mutually perpendicular with equal magnitudes

∣ a ∣ = ∣ b ∣ = ∣ c ∣

:

Expanding each term and using orthogonality:

∣ a ∣^{2} (r - b) - (a \cdot r) a + ∣ b ∣^{2} (r - c) - (b \cdot r) b + ∣ c ∣^{2} (r - a) - (c \cdot r) c = 0

Simplifying with

∣ a ∣^{2} = ∣ b ∣^{2} = ∣ c ∣^{2}

∣ a ∣^{2} (3 r - (a + b + c)) - ((a \cdot r) a + (b \cdot r) b + (c \cdot r) c) = 0

Letting

r = x a + y b + z c

and solving gives

r = \frac{1}{2} (a + b + c)

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option