JEE Mathematics 2026 Question with Solution

Given: $\overrightarrow{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k}$, $\overrightarrow{AD} = \hat{i} + 2\hat{j} + \lambda \hat{k}$, and $\vec{v} = \hat{i} + \hat{j} + \hat{k}$.

Find: The value of $2\alpha - \beta$, where $\alpha > \beta$ are the roots of $\lambda^2 x^2 - 6\lambda x + 5 = 0$.

The diagonal of parallelogram $ABCD$ is

$$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}$$

So,

$$\overrightarrow{AC} = (2+1)\hat{i} + (4+2)\hat{j} + (-5+\lambda)\hat{k} = 3\hat{i} + 6\hat{j} + (\lambda-5)\hat{k}$$

The magnitude of projection of $\vec{v}$ on $\overrightarrow{AC}$ is

$$\left|\text{proj}_{AC}\vec{v}\right| = \frac{|\vec{v} \cdot \overrightarrow{AC}|}{|\overrightarrow{AC}|}$$

Given this length is $1$, therefore

$$\frac{|\vec{v} \cdot \overrightarrow{AC}|}{|\overrightarrow{AC}|} = 1$$

Now,

$$\vec{v} \cdot \overrightarrow{AC} = (1)(3) + (1)(6) + (1)(\lambda-5) = \lambda + 4$$

and

$$|\overrightarrow{AC}| = \sqrt{3^2 + 6^2 + (\lambda-5)^2} = \sqrt{45 + (\lambda-5)^2}$$

Hence,

$$\frac{|\lambda + 4|}{\sqrt{45 + (\lambda-5)^2}} = 1$$

Squaring both sides,

$$(\lambda + 4)^2 = 45 + (\lambda - 5)^2$$

Expanding,

$$\lambda^2 + 8\lambda + 16 = \lambda^2 - 10\lambda + 25 + 45$$

So,

$$18\lambda = 54 \Rightarrow \lambda = 3$$

Now the quadratic becomes

$$9x^2 - 18x + 5 = 0$$

Using the quadratic formula,

$$x = \frac{18 \pm \sqrt{324 - 180}}{18} = \frac{18 \pm 12}{18}$$

Thus,

$$\alpha = \frac{5}{3}, \quad \beta = \frac{1}{3}$$

Therefore,

$$2\alpha - \beta = 2\left(\frac{5}{3}\right) - \frac{1}{3} = \frac{9}{3} = 3$$

Therefore, the correct option is C.

Given: The projection of $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ on diagonal $\overrightarrow{AC}$ has magnitude $1$.

Find: First determine $\lambda$, then evaluate $2\alpha - \beta$.

Since $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}$,

$$\overrightarrow{AC} = 3\hat{i} + 6\hat{j} + (\lambda - 5)\hat{k}$$

For projection magnitude,

$$\left|\text{proj}_{\overrightarrow{AC}} \vec{v}\right| = \frac{|\vec{v} \cdot \overrightarrow{AC}|}{|\overrightarrow{AC}|} = 1$$

Compute the numerator:

$$\vec{v} \cdot \overrightarrow{AC} = 3 + 6 + (\lambda - 5) = \lambda + 4$$

Compute the denominator:

$$|\overrightarrow{AC}| = \sqrt{9 + 36 + (\lambda - 5)^2} = \sqrt{45 + (\lambda - 5)^2}$$

So,

$$|\lambda + 4| = \sqrt{45 + (\lambda - 5)^2}$$

After squaring,

$$(\lambda + 4)^2 = 45 + (\lambda - 5)^2$$

This gives

$$\lambda^2 + 8\lambda + 16 = 45 + \lambda^2 - 10\lambda + 25$$

Hence,

$$18\lambda = 54$$

and so

$$\lambda = 3$$

Substitute into the given equation:

$$\lambda^2 x^2 - 6\lambda x + 5 = 0 \Rightarrow 9x^2 - 18x + 5 = 0$$

Solving,

$$x = \frac{18 \pm 12}{18}$$

Therefore the roots are

$$\frac{30}{18} = \frac{5}{3}, \quad \frac{6}{18} = \frac{1}{3}$$

Since $\alpha > \beta$,

$$\alpha = \frac{5}{3}, \quad \beta = \frac{1}{3}$$

Now evaluate

$$2\alpha - \beta = \frac{10}{3} - \frac{1}{3} = \frac{9}{3} = 3$$

Therefore, the required value is $3$.