-\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 $

Vector Algebra

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