JEE Mathematics 2023 Question with Solution

Given: $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$.

Also,

$$\vec{a} \cdot \vec{c} = 11$$ $$\vec{b} \cdot (\vec{a} \times \vec{c}) = 27$$

and

$$\vec{b} \cdot \vec{c} = -\sqrt{3}|\vec{b}|$$

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

From the solution working,

$$\vec{a} \cdot \vec{b} = 0$$

so $\vec{a}$ and $\vec{b}$ are perpendicular.

The solution further uses

$$\left|\vec{b} \times (\vec{a} \times \vec{c})\right| = 3\sqrt{14}$$

and

$$\left|\vec{b} \cdot (\vec{a} \times \vec{c})\right| = 27$$

Let $\theta$ be the angle between $\vec{b}$ and $\vec{a} \times \vec{c}$. Then

$$\left|\vec{b} \times (\vec{a} \times \vec{c})\right| = |\vec{b}|\,|\vec{a} \times \vec{c}|\sin\theta = 3\sqrt{14}$$

and

$$\left|\vec{b} \cdot (\vec{a} \times \vec{c})\right| = |\vec{b}|\,|\vec{a} \times \vec{c}|\cos\theta = 27$$

Dividing the two relations,

$$\tan\theta = \frac{3\sqrt{14}}{27} = \frac{\sqrt{14}}{9}$$

which gives

$$\sin\theta = \frac{\sqrt{14}}{\sqrt{95}}$$

Using this in the magnitude relation stated in the solution,

$$\left|\vec{b} \times (\vec{a} \times \vec{c})\right| = 3\sqrt{95}$$

Therefore,

$$|\vec{a} \times \vec{c}|^2 = 3 \times 95 = 285$$

Therefore, the required numerical value is $285$.

Given: the scalar product and mixed product conditions involving $\vec{a}, \vec{b}, \vec{c}$.

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

The extracted solution resolves the problem by treating $\vec{a} \times \vec{c}$ as a vector making angle $\theta$ with $\vec{b}$. Then:

$$|\vec{b}|\,|\vec{a} \times \vec{c}|\cos\theta = 27$$

and the perpendicular component gives

$$|\vec{b}|\,|\vec{a} \times \vec{c}|\sin\theta = 3\sqrt{14}$$

Hence,

$$\sin\theta = \frac{\sqrt{14}}{\sqrt{95}}$$

so the common magnitude factor is evaluated in the solution as

$$|\vec{b}|\,|\vec{a} \times \vec{c}| = 3\sqrt{95}$$

Finally, the extracted working concludes

$$|\vec{a} \times \vec{c}|^2 = 285$$

Thus the answer is $285$.