JEE Mathematics 2023 Question with Solution

Given: $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$, $\vec{b} = \hat{i} - 2\hat{j} - 2\hat{k}$, $\vec{c} = -\hat{i} + 4\hat{j} + 3\hat{k}$, and $\vec{a} \cdot \vec{d} = 18$.

Find: $|\vec{a} \times \vec{d}|^2$.

Since $\vec{d}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, it must be parallel to $\vec{b} \times \vec{c}$.

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

Using $\vec{a} \cdot \vec{d} = 18$,

$$(2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot \lambda (2\hat{i} - \hat{j} + 2\hat{k}) = 18$$ $$\lambda (4 - 3 + 8) = 18$$ $$9\lambda = 18 \Rightarrow \lambda = 2$$

Therefore,

$$\vec{d} = 2(2\hat{i} - \hat{j} + 2\hat{k}) = 4\hat{i} - 2\hat{j} + 4\hat{k}$$

Now use the identity

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

Compute the magnitudes:

$$|\vec{a}|^2 = 2^2 + 3^2 + 4^2 = 29$$ $$|\vec{d}|^2 = 4^2 + (-2)^2 + 4^2 = 36$$

Hence,

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

Therefore, the correct option is C and the value is $720$.

Given: $\vec{d}$ is perpendicular to both $\vec{b}$ and $\vec{c}$.

Find: $|\vec{a} \times \vec{d}|^2$.

Because a vector perpendicular to both vectors is along their cross product,

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

Now expand

$$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -2 \\ -1 & 4 & 3 \end{vmatrix}$$ $$= (-6 + 8)\hat{i} - (3 - 2)\hat{j} + (4 - 2)\hat{k}$$ $$= 2\hat{i} - \hat{j} + 2\hat{k}$$

So,

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

Using the given dot product,

$$(2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot \lambda (2\hat{i} - \hat{j} + 2\hat{k}) = 18$$ $$\lambda (4 - 3 + 8) = 18$$ $$9\lambda = 18 \Rightarrow \lambda = 2$$

Thus,

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

Now apply

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

with

$$|\vec{a}|^2 = 29, \qquad |\vec{d}|^2 = 36, \qquad \vec{a} \cdot \vec{d} = 18$$

Therefore,

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

So the required value is $720$.