Let the vectors $\overrightarrow{\mathrm{u}}_{1}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{a} \hat{\mathrm{k}}$, $\overrightarrow{\mathrm{u}}_{2}=\hat{\mathrm{i}}+\mathrm{b} \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{u}}_{3}=\mathrm{c} \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ be coplanar. If the vectors $\overrightarrow{\mathrm{v}}_{1}=(\mathrm{a}+\mathrm{b}) \hat{\mathrm{i}}+\mathrm{c} \hat{\mathrm{j}}+\mathrm{c} \hat{\mathrm{k}}$, $\overrightarrow{\mathrm{v}}_{2}=\mathrm{a} \hat{\mathrm{i}}+(\mathrm{b}+\mathrm{c}) \hat{\mathrm{j}}+\mathrm{a} \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{v}}_{3}=\mathrm{b} \hat{\mathrm{i}}+\mathrm{b} \hat{\mathrm{j}}+(\mathrm{c}+\mathrm{a}) \hat{\mathrm{k}}$ are also coplanar, then $6(\mathrm{a}+\mathrm{b}+\mathrm{c})$ is equal to

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Vector Algebra

Let the vectors

u_{1} = \hat{i} + \hat{j} + a \hat{k}

u_{2} = \hat{i} + b \hat{j} + \hat{k}

and

u_{3} = c \hat{i} + \hat{j} + \hat{k}

be coplanar. If the vectors

v_{1} = (a + b) \hat{i} + c \hat{j} + c \hat{k}

v_{2} = a \hat{i} + (b + c) \hat{j} + a \hat{k}

and

v_{3} = b \hat{i} + b \hat{j} + (c + a) \hat{k}

are also coplanar, then

6 (a + b + c)

is equal to

Answer: (C)

Coplanarity of

u_{1}, u_{2}, u_{3}

requires

11 c 1 b 1 a 11 = 0.

Expanding,

1 (b - 1) - 1 (1 - c) + a (1 - b c) = a + b + c - ab c - 2 = 0

, so

a + b + c = 2 + ab c . (1)

Coplanarity of

v_{1}, v_{2}, v_{3}

requires

a + b a b c b + c b c a c + a = 0.

Expanding this determinant gives

4 ab c = 0

, so

ab c = 0. (2)

Substituting

(2)

into

(1)

yields

a + b + c = 2

, and hence

6 (a + b + c) = 12

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option