Let c→ be a vector perpendicular to the vectors a→=i^+j^-k^ and b→=i^+2j^+k^. If c→·(i^+j^+3k^)=8, then the value of c→·(a→×b→) is equal to

Correct answer: 28. c→=λ(a→×b→) a→×b→= i^ j^ k^ 1 1 -1 1 2 1 (a→×b→)=3i^-2j^+k^ c→·(i^+j^+3k^)=λ(3i^-2j^+k^)·(i^+j^+3k^) ⇒λ(4)=8⇒λ=2 c→=2(a→×b→) Hence, c→·(a→×b→)=28

Q59 / 90 · JEE Main 2021 (16 Mar Shift 2) · Mathematics

Let $\vec{c}$ be a vector perpendicular to the vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + \hat{k}$ . If $\vec{c} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 8$ , then the value of $\vec{c} \cdot (\vec{a} \times \vec{b})$ is equal to

Correct answer: 28

Solution

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

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

$(\vec{a} \times \vec{b}) = 3 \hat{i} - 2 \hat{j} + \hat{k}$

$\vec{c} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = λ (3 \hat{i} - 2 \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j} + 3 \hat{k})$

$\Rightarrow λ (4) = 8 \Rightarrow λ = 2$

$\vec{c} = 2 (\vec{a} \times \vec{b})$

Hence, $\vec{c} \cdot (\vec{a} \times \vec{b}) = 28$

Hence, the correct answer is 28.

Let c→ be a vector perpendicular to the vectors a→=i^+j^-k^ and b→=i^+2j^+k^. If c→·(i^+j^+3k^)=8, then the value of c→·(a→×b→) is equal to

Solution

Relevant Questions

Let $\vec{c}$ be a vector perpendicular to the vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + \hat{k}$ . If $\vec{c} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 8$ , then the value of $\vec{c} \cdot (\vec{a} \times \vec{b})$ is equal to