Let a→=i^+j^+2k^ and b→=-i^+2j^+3k^. Then the vector product a→+b→×a→×a→-b→×b→×b→ is equal to :

Correct answer: (B) 734i^-5j^+3k^. a→=i^+j^+2k^ b→=-i^+2j^+3k^ a→+b→=3j^+5k^;a→·b→=-1+2+6=7 Now a→×a→-b→×b→×b→ =a→×a→×b→-b→×b→×b→ =a→×a→×b→-0×b→ =a→×a→×b→×b→ =a→·b→a→-a→·a→b→×b→ =a→·b→a→×b→-a→·a→b→×b→ =a→·b→a→×b→ Here, a→×b→= i^ j^ k^ 1 1 2 -1 2 3 =-i^-5j^+3k^ ∴a→·b→a→×b→=7-i^-5j^+3k^ So, a→+b→×7-i^-5j^+3k^ 70i^+3j^+5k^×-i^-5j^+3k^ =7 i^ j^ k^ 0 3 5 -1 -5 3 =734i^-5j^+3k^ =734i^-5j^+3k^

Q33 / 90 · JEE Main 2021 (27 Jul Shift 1) · Mathematics

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 :

A
$5 (34 \hat{i} - 5 \hat{j} + 3 \hat{k})$
Correct answer
B
$7 (34 \hat{i} - 5 \hat{j} + 3 \hat{k})$
C
$7 (30 \hat{i} - 5 \hat{j} + 7 \hat{k})$
D
$5 (30 \hat{i} - 5 \hat{j} + 7 \hat{k})$

Solution

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

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

$\vec{a} + \vec{b} = 3 \hat{j} + 5 \hat{k}; \vec{a} \cdot \vec{b} = - 1 + 2 + 6 = 7$

Now $((\vec{a} \times ((\vec{a} - \vec{b}) \times \vec{b})) \times \vec{b})$

$= ((\vec{a} \times (\vec{a} \times \vec{b} - \vec{b} \times \vec{b})) \times \vec{b})$

$= (\vec{a} \times (\vec{a} \times \vec{b} - 0)) \times \vec{b}$

$= (\vec{a} \times (\vec{a} \times \vec{b})) \times \vec{b}$

$= ((\vec{a} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{b}) \times \vec{b}$

$= (\vec{a} \cdot \vec{b}) (\vec{a} \times \vec{b}) - (\vec{a} \cdot \vec{a}) (\vec{b} \times \vec{b})$

$= (\vec{a} \cdot \vec{b}) (\vec{a} \times \vec{b})$

Here, $\vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ - 1 & 2 & 3 \end{matrix}| = - \hat{i} - 5 \hat{j} + 3 \hat{k}$

$∴ (\vec{a} \cdot \vec{b}) (\vec{a} \times \vec{b}) = 7 (- \hat{i} - 5 \hat{j} + 3 \hat{k})$

So, $(\vec{a} + \vec{b}) \times (7 (- \hat{i} - 5 \hat{j} + 3 \hat{k}))$

$7 (0 \hat{i} + 3 \hat{j} + 5 \hat{k}) \times (- \hat{i} - 5 \hat{j} + 3 \hat{k})$

$= 7 |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 5 \\ - 1 & - 5 & 3 \end{matrix}|$

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

$= 7 [34 \hat{i} - 5 \hat{j} + 3 \hat{k}]$

Hence, the correct answer is option: 'B'.

Let a→=i^+j^+2k^ and b→=-i^+2j^+3k^. Then the vector product a→+b→×a→×a→-b→×b→×b→ is equal to :

Solution

Relevant Questions

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 :