Let $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=-\hat{i}+2 \hat{j}+3 \hat{k}$. Then the vector product $(\vec{a}+\vec{b}) \times((\vec{a} \times((\vec{a}-\vec{b}) \times \vec{b})) \times \vec{b})$ is equal to :

(D) $7(34 \hat{i}-5 \hat{j}+3 \hat{k})$

(A) $5(34 \hat{i}-5 \hat{j}+3 \hat{k})$

(B) $5(30 \hat{i}-5 \hat{j}+7 \hat{k})$

(C) $7(30 \hat{i}-5 \hat{j}+7 \hat{k})$

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Vector Algebra

Let

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

and

b = - \hat{i} + 2 \hat{j} + 3 \hat{k}

. Then the vector product

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

is equal to :

Answer: (D)

Here

a \cdot b = - 1 + 2 + 6 = 7

a \cdot a = 6

a + b = 3 \hat{j} + 5 \hat{k}

and

a \times b = - \hat{i} - 5 \hat{j} + 3 \hat{k}

Because

b \times b = 0

(a - b) \times b = a \times b

. The triple product expansion then gives

a \times (a \times b) = (a \cdot b) a - (a \cdot a) b = 7 a - 6 b

and crossing this with

b

removes the second term:

(7 a - 6 b) \times b = 7 (a \times b)

The required product is therefore

7 (a + b) \times (a \times b) = 7 \hat{i} 0 - 1 \hat{j} 3 - 5 \hat{k} 53 = 7 (34 \hat{i} - 5 \hat{j} + 3 \hat{k})

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option