JEE Mathematics 2023 Question with Solution

Given: $\vec{P} = 3 \hat{i} + \sqrt{3} \hat{j} + 2 \hat{k}$ and $\vec{Q} = 4 \hat{i} + \sqrt{3} \hat{j} + 2.5 \hat{k}$.

Find: The value of $x$ in the unit vector expression $\frac{1}{x}\left(\sqrt{3} \hat{i} + \hat{j} - 2\sqrt{3} \hat{k}\right)$.

First compute the cross product:

$$\vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & \sqrt{3} & 2 \\ 4 & \sqrt{3} & 2.5 \end{vmatrix}$$

Expanding the determinant,

$$\vec{P} \times \vec{Q} = \hat{i}\begin{vmatrix} \sqrt{3} & 2 \\ \sqrt{3} & 2.5 \end{vmatrix} - \hat{j}\begin{vmatrix} 3 & 2 \\ 4 & 2.5 \end{vmatrix} + \hat{k}\begin{vmatrix} 3 & \sqrt{3} \\ 4 & \sqrt{3} \end{vmatrix}$$

Now evaluate each minor:

$$\hat{i}: (\sqrt{3})(2.5) - (\sqrt{3})(2) = 0.5\sqrt{3}$$ $$\hat{j}: (3)(2.5) - (4)(2) = 7.5 - 8 = -0.5$$ $$\hat{k}: (3)(\sqrt{3}) - (4)(\sqrt{3}) = -\sqrt{3}$$

So,

$$\vec{P} \times \vec{Q} = 0.5\sqrt{3} \, \hat{i} + 0.5 \, \hat{j} - \sqrt{3} \, \hat{k}$$

Its magnitude is

$$|\vec{P} \times \vec{Q}| = \sqrt{\left(0.5\sqrt{3}\right)^2 + (0.5)^2 + \left(-\sqrt{3}\right)^2}$$ $$|\vec{P} \times \vec{Q}| = \sqrt{\frac{3}{4} + \frac{1}{4} + 3} = \sqrt{4} = 2$$

Hence the unit vector is

$$\frac{\vec{P} \times \vec{Q}}{|\vec{P} \times \vec{Q}|} = \frac{1}{2}\left(\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} - \sqrt{3} \hat{k}\right)$$

which simplifies to

$$\frac{1}{4}\left(\sqrt{3} \hat{i} + \hat{j} - 2\sqrt{3} \hat{k}\right)$$

Comparing with the given form $\frac{1}{x}\left(\sqrt{3} \hat{i} + \hat{j} - 2\sqrt{3} \hat{k}\right)$, we get $x = 4$.

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

Given: The direction vector is obtained from $\vec{P} \times \vec{Q}$.

Find: The scalar $x$.

From the cross product working,

$$\vec{P} \times \vec{Q} = 0.5\sqrt{3} \, \hat{i} + 0.5 \, \hat{j} - \sqrt{3} \, \hat{k}$$

Divide by its magnitude $2$ to get the unit vector:

$$\text{Unit vector} = \frac{1}{2}\left(0.5\sqrt{3} \, \hat{i} + 0.5 \, \hat{j} - \sqrt{3} \, \hat{k}\right)$$ $$\text{Unit vector} = \frac{\sqrt{3}}{4} \hat{i} + \frac{1}{4} \hat{j} - \frac{\sqrt{3}}{2} \hat{k}$$

Factor out $\frac{1}{4}$:

$$\text{Unit vector} = \frac{1}{4}\left(\sqrt{3} \hat{i} + \hat{j} - 2\sqrt{3} \hat{k}\right)$$

Now compare this directly with the given expression:

$$\frac{1}{x}\left(\sqrt{3} \hat{i} + \hat{j} - 2\sqrt{3} \hat{k}\right)$$

Thus,

$$\frac{1}{x} = \frac{1}{4}$$

So,

$$x = 4$$

Therefore, the required numerical value is $4$.