QJEEM24Ph67a5: JEE Main 2024 Physics - Class 11 JEEM24Ph67a5 Solution: For three vectors A → = ( − x i ^ − 6 j ^ − 2 k ^ ) , B → = ( − …

To determine the value of $x$ , given the vectors $\vec{A}=(-x \hat{i} - 6 \hat{j} - 2 \hat{k})$ , $\vec{B}=(-\hat{i} + 4 \hat{j} + 3 \hat{k})$ , and $\vec{C}=(-8 \hat{i} - \hat{j} + 3 \hat{k})$ , and the condition $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$ , we proceed as follows:

First, we calculate the cross product $\vec{B} \times \vec{C}$ :

$\vec{B} \times \vec{C} = 15 \hat{i} - 21 \hat{j} + 33 \hat{k}$

Next, using the condition $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$ , we compute the dot product:

$\vec{A} \cdot (\vec{B} \times \vec{C}) = (-x \hat{i} - 6 \hat{j} - 2 \hat{k}) \cdot (15 \hat{i} - 21 \hat{j} + 33 \hat{k})$

$\Rightarrow (-x)(15) + (-6)(-21) + (-2)(33) = 0$

$\Rightarrow -15x + 126 - 66 = 0$

Solving this equation for $x$ :

$-15x + 60 = 0$

$\Rightarrow x = 4$

Thus, the value of $x$ is $4$ .