JEE Mathematics 2023 Question with Solution

Given: $\vec{v} = \alpha \hat{i} + 2 \hat{j} - 3 \hat{k}$, $\vec{w} = 2\alpha \hat{i} + \hat{j} - \hat{k}$, and $|\hat{u}| = \alpha > 0$.

Find: $m+n$, where $|\hat{u} \cdot \hat{i}|^2 = \frac{m}{n}$.

Using the scalar triple product,

$$[\vec{u}\vec{v}\vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$$

Its minimum value is

$$\min\left(|u|\,|\vec{v} \times \vec{w}|\cos\theta\right) = -\alpha\sqrt{3401}$$

So,

$$\cos\theta = -1$$

and

$$|u| = \alpha \text{ (Given)}$$

Hence,

$$|\vec{v} \times \vec{w}| = \sqrt{3401}$$

Now compute the cross product:

$$\vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \alpha & 2 & -3 \\ 2\alpha & 1 & -1 \end{vmatrix}$$

Therefore,

$$\vec{v} \times \vec{w} = \hat{i} - 5\alpha \hat{j} - 3\alpha \hat{k}$$

So its magnitude is

$$|\vec{v} \times \vec{w}| = \sqrt{1+25\alpha^{2}+9\alpha^{2}} = \sqrt{3401}$$

Thus,

$$34\alpha^{2}=3400$$ $$\alpha^{2}=100$$

Since $\alpha>0$,

$$\alpha=10$$

Because the minimum occurs when $\vec{u}$ is opposite to $\vec{v}\times\vec{w}$, we take

$$\vec{u}=\lambda(\hat{i}-5\alpha\hat{j}-3\alpha\hat{k})$$

Using $|\vec{u}|=\alpha$,

$$\alpha^{2}=\lambda^{2}(1+25\alpha^{2}+9\alpha^{2})$$

Substituting $\alpha^{2}=100$,

$$100=\lambda^{2}(1+34\times100)$$

So,

$$\lambda^{2}=\frac{100}{3401} = \frac{m}{n}$$

Hence $m=100$ and $n=3401$, so

$$m+n=3501$$

Therefore, the required answer is $3501$.

Given: the minimum scalar triple product is $-\alpha\sqrt{3401}$.

Find: $m+n$.

Expand

$$\vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \alpha & 2 & -3\\ 2\alpha & 1 & -1 \end{vmatrix}$$

This gives

$$\vec{v} \times \vec{w} = \hat{i}(-2+3)-\hat{j}(-\alpha+6\alpha)+\hat{k}(\alpha-4\alpha)$$ $$= \hat{i}-5\alpha\hat{j}-3\alpha\hat{k}$$

Therefore,

$$|\vec{v} \times \vec{w}| = \sqrt{1+25\alpha^2+9\alpha^2}$$

Given this equals $\sqrt{3401}$,

$$1+34\alpha^2 = 3401$$ $$34\alpha^2 = 3400$$ $$\alpha^2 = 100$$

Now $\vec{u}$ must be parallel or antiparallel to $\vec{v}\times\vec{w}$ for the triple product to attain an extreme value. Since the minimum is negative, it is antiparallel. Thus the coefficient of $\hat{i}$ in $\vec{u}$ is $\lambda$, so

$$|\hat{u}\cdot\hat{i}|^2 = \lambda^2$$

From $|\hat{u}|=\alpha$,

$$\lambda^2(1+25\alpha^2+9\alpha^2)=\alpha^2$$ $$\lambda^2 = \frac{\alpha^2}{1+34\alpha^2} = \frac{100}{3401}$$

Hence,

$$\frac{m}{n} = \frac{100}{3401}$$

So,

$$m+n = 100+3401 = 3501$$

Therefore, the answer is $3501$.