QJEEM23Maee7d: JEE Main 2023 Physics - Class 11 JEEM23Maee7d Solution: Let and . If is a vector perpendicular to both and , and , then …

Given vectors :

$\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$

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

$\vec{c} = -\hat{i} + 4\hat{j} + 3\hat{k}$

Since $\vec{d}$ is perpendicular to both $\vec{b}$ and $\vec{c}$ , its direction is given by their cross product :

$\vec{d} = \lambda(\vec{b} \times \vec{c})$

$\begin{aligned}\vec{b} \times \vec{c} & =\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\1 & -2 & -2 \\-1 & 4 & 3\end{array}\right| \\\\& =\hat{\mathbf{i}}(-6+8)-\hat{\mathbf{j}}(3-2)+\hat{\mathbf{k}}(4-2)=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\end{aligned}$

Thus, $\vec{b} \times \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k}$

Given this, $\vec{d}$ can be expressed as :

$\vec{d} = \lambda(2\hat{i} - \hat{j} + 2\hat{k})$

Using the given condition that $\vec{a} \cdot \vec{d} = 18$ :

$(2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot \lambda(2\hat{i} - \hat{j} + 2\hat{k}) = 18$

$\Rightarrow 2\lambda(2) - 3\lambda(1) + 4\lambda(2) = 18$

$\Rightarrow \lambda(4 + 8 - 3) = 18$

$\Rightarrow 9\lambda = 18$

$\Rightarrow \lambda = 2$

So, $\vec{d} = 4\hat{i} - 2\hat{j} + 4\hat{k}$

Now, using the identity :

$|\vec{a} \times \vec{d}|^2 = |\vec{a}|^2 |\vec{d}|^2 - (\vec{a} \cdot \vec{d})^2$

Given $|\vec{a}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$

and $|\vec{d}| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$

Using your corrected calculations :

$|\vec{a} \times \vec{d}|^2 = 29 \times 36 - 18^2 = 1044 - 324 = 720$