JEE Mathematics 2024 Question with Solution

Given: $\vec{a} = \hat{i} + \alpha \hat{j} + \beta \hat{k}$, the angle between $\vec{a}$ and $\vec{b}$ is $\pi/4$, $|\vec{b}| = 6$, and $|\vec{a} \times \vec{b}| = 3\sqrt{2}$.

Find: The value of $(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2$.

Use the cross product magnitude formula:

$$|\vec{a} \times \vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta$$

So,

$$3\sqrt{2} = |\vec{a}| \times 6 \times \frac{\sqrt{2}}{2}$$

which gives

$$|\vec{a}| = 1$$

Now,

$$|\vec{a}|^2 = 1 + \alpha^2 + \beta^2$$

Hence,

$$1 + \alpha^2 + \beta^2 = 1$$

so

$$\alpha^2 + \beta^2 = 5$$

Also,

$$|\vec{a} \times \vec{b}| = 1 \times 6 \times \frac{\sqrt{2}}{2} = 3\sqrt{2}$$

Therefore,

$$(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2 = 5 \times 18 = 90$$

Therefore, the correct option is A.

Given: $\vec{a} = \hat{i} + \alpha \hat{j} + \beta \hat{k}$, $\theta = \pi/4$, $|\vec{b}| = 6$, and $|\vec{a} \times \vec{b}| = 3\sqrt{2}$.

Find: The value of $(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2$.

From

$$|\vec{a} \times \vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta$$

and

$$\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

we get

$$3\sqrt{2} = |\vec{a}| \cdot 6 \cdot \frac{\sqrt{2}}{2}$$

Thus,

$$3\sqrt{2} = 3\sqrt{2}\,|\vec{a}|$$

so

$$|\vec{a}| = 1$$

Now the vector $\vec{a} = \hat{i} + \alpha \hat{j} + \beta \hat{k}$ has magnitude

$$|\vec{a}| = \sqrt{1 + \alpha^2 + \beta^2}$$

Therefore,

$$\sqrt{1 + \alpha^2 + \beta^2} = 1$$

Squaring both sides,

$$1 + \alpha^2 + \beta^2 = 6$$

Hence,

$$\alpha^2 + \beta^2 = 5$$

Now,

$$|\vec{a} \times \vec{b}|^2 = (3\sqrt{2})^2 = 18$$

So,

$$(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2 = 5 \times 18 = 90$$

Therefore, the value is $90$, so the correct option is A.