Let a vector \(\vec{a}\) be coplanar with vectors \(\vec{b}=2 \hat{i}+\hat{j}+\hat{k}\) and \(\vec{c}=\hat{i}-\hat{j}+\hat{k}\). If \(\vec{a}\) is perpendicular to \(\vec{d}=3 \hat{i}+2 \hat{j}+6 \hat{k}\) and \(|\vec{a}|=\sqrt{10}\). Then a possible value of \([\vec{a} \vec{b} \vec{c}]+[\vec{a} \vec{b} \vec{d}]+[\vec{a} \vec{c} \vec{d}]\) is equal to:

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Vector Algebra

Let a vector

a

be coplanar with vectors

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

and

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

. If

a

is perpendicular to

d = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}

and

∣ a ∣ = 10

. Then a possible value of

[a b c] + [a b d] + [a c d]

is equal to:

Answer: (C)

Coplanarity with

b

and

c

allows

a = x b + y c

. With

b \cdot d = 14

and

c \cdot d = 7

, the condition

a \cdot d = 0

gives

14 x + 7 y = 0

, so

y = - 2 x

and

a = x (2 \hat{i} + \hat{j} + \hat{k}) - 2 x (\hat{i} - \hat{j} + \hat{k}) = x (3 \hat{j} - \hat{k})

Then

∣ a ∣ = ∣ x ∣ 10 = 10

gives

x = \pm 1

. Since

a

lies in the plane of

b

and

c

[a b c] = 0

. Also

b \times d = 4 \hat{i} - 9 \hat{j} + \hat{k}

and

c \times d = - 8 \hat{i} - 3 \hat{j} + 5 \hat{k}

, so for

a = 3 \hat{j} - \hat{k}

[a b d] = - 27 - 1 = - 28, [a c d] = - 9 - 5 = - 14

A possible value of the sum of the three triple products is

- 42

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option