JEE Mathematics 2025 Question with Solution

Given: $\vec{u} = 3\hat{i} - \hat{j}$ and $\vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k}$ with angle between them equal to $\cos^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}\right)$.

Find: $|\vec{v}_1|^2 + |\vec{v}_2|^2$ where $\vec{v}_1 \parallel \vec{u}$ and $\vec{v}_2 \perp \vec{u}$.

Using the angle formula,

$$\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}|\,|\vec{v}|}$$

we first compute the dot product and magnitudes.

$$\vec{u} \cdot \vec{v} = 3 \cdot 2 + (-1) \cdot 1 = 5$$$$|\vec{u}| = \sqrt{3^2 + (-1)^2} = \sqrt{10}$$$$|\vec{v}| = \sqrt{2^2 + 1^2 + \lambda^2} = \sqrt{5 + \lambda^2}$$

Now,

$$\frac{5}{\sqrt{10}\,\sqrt{5 + \lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}}$$

Squaring and simplifying gives

$$\frac{25}{10(5 + \lambda^2)} = \frac{5}{28}$$$$\frac{5}{2(5 + \lambda^2)} = \frac{5}{28}$$$$2(5 + \lambda^2) = 28$$$$\lambda^2 = 9$$

Since $\lambda > 0$,

$$\lambda = 3$$

For the orthogonal decomposition $\vec{v} = \vec{v}_1 + \vec{v}_2$ with $\vec{v}_1 \perp \vec{v}_2$, we use

$$|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2$$

Therefore,

$$|\vec{v}_1|^2 + |\vec{v}_2|^2 = |\vec{v}|^2 = 5 + \lambda^2 = 5 + 9 = 14$$

So, the correct option is B.

Given: $\vec{v} = \vec{v}_1 + \vec{v}_2$ where $\vec{v}_1$ is parallel to $\vec{u}$ and $\vec{v}_2$ is perpendicular to $\vec{u}$.

Find: The value of $|\vec{v}_1|^2 + |\vec{v}_2|^2$.

Because one component is along $\vec{u}$ and the other is perpendicular to $\vec{u}$, the vectors $\vec{v}_1$ and $\vec{v}_2$ are mutually perpendicular. Hence Pythagoras applies directly:

$$|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2$$

So the problem reduces to finding $|\vec{v}|^2$.

From the angle condition,

$$\frac{\vec{u} \cdot \vec{v}}{|\vec{u}|\,|\vec{v}|} = \frac{\sqrt{5}}{2\sqrt{7}}$$

with

$$\vec{u} \cdot \vec{v} = 5, \qquad |\vec{u}| = \sqrt{10}, \qquad |\vec{v}| = \sqrt{5 + \lambda^2}$$

Thus,

$$\frac{5}{\sqrt{10}\sqrt{5 + \lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}}$$

which gives $\lambda^2 = 9$ and hence $\lambda = 3$.

Therefore,

$$|\vec{v}|^2 = 5 + \lambda^2 = 14$$

So, $|\vec{v}_1|^2 + |\vec{v}_2|^2 = 14$, and the correct option is B.