QJEEM25Mab1da: JEE Main 2025 Mathematics - Class 12 JEEM25Mab1da Solution: Let and , be two lines, which intersect at the point . If is…

Explanation

To find the value of $26 \alpha(\mathrm{PB})^2$ , we proceed as follows:

Intersection Point $B$ of $\mathrm{L}_1$ and $\mathrm{L}_2$

We are given the equations of the lines:

$\mathrm{L}_1 : \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0}$

$\mathrm{L}_2 : \frac{x-2}{2} = \frac{y}{0} = \frac{z+4}{\alpha}$

For $\mathrm{L}_1$ , $z+1 = 0$ , which implies $z = -1$ .

For $\mathrm{L}_2$ , since $y = 0$ , the direction ratios can be matched using the parameter $\mu$ :

$\begin{aligned} & x = 3\lambda + 1, \quad y = -\lambda + 1, \quad z = -1, \\ & x = 2\mu + 2, \quad y = 0, \quad z = \alpha \mu - 4. \end{aligned}$

Setting the coordinates equal for intersection:

$\begin{aligned} & 3\lambda + 1 = 2\mu + 2, \\ & -\lambda + 1 = 0, \\ & -1 = \alpha \mu - 4. \end{aligned}$

From $-\lambda + 1 = 0$ , we find $\lambda = 1$ .

Substituting $\lambda = 1$ into $3\lambda + 1 = 2\mu + 2$ gives:

$3(1) + 1 = 2\mu + 2 \implies \mu = 1.$

Also, substituting $\mu = 1$ into $-1 = \alpha \mu - 4$ gives:

$-1 = \alpha(1) - 4 \implies \alpha = 3.$

So, the intersection point $B$ is:

$B(4, 0, -1).$

Foot of Perpendicular from $A$ to $\mathrm{L}_2$

The point $P$ on $\mathrm{L}_2$ is given by:

$P(2\delta + 2, 0, 3\delta - 4).$

For vector $\overrightarrow{AP} = (2\delta + 1, -1, 3\delta - 3)$ , since $AP$ is perpendicular to $\mathrm{L}_2$ , the dot product should be zero:

$\begin{aligned} (2\delta + 1)\cdot 2 + (-1)\cdot 0 + (3\delta - 3)\cdot 3 &= 0. \end{aligned}$

Simplifying gives:

$4\delta + 2 + 9\delta - 9 = 0 \implies 13\delta = 7 \implies \delta = \frac{7}{13}.$

So, the coordinates of point $P$ are:

$P\left(\frac{40}{13}, 0, \frac{-31}{13}\right).$

Distance $\mathrm{PB}$ and Calculation

The vector $\overrightarrow{PB}$ is:

$\overrightarrow{PB} = \left(4 - \frac{40}{13}, 0 - 0, -1 - \left(\frac{-31}{13}\right)\right).$

Calculating the components:

$\overrightarrow{PB} = \left(\frac{12}{13}, 0, \frac{18}{13}\right).$

The square of the distance $(PB)^2$ is:

$\left(\frac{12}{13}\right)^2 + \left(0\right)^2 + \left(\frac{18}{13}\right)^2 = \frac{144}{169} + \frac{324}{169} = \frac{468}{169}.$

Thus, $26 \alpha (PB)^2$ is:

$26 \times 3 \times \frac{468}{169} = 216.$

So, the final value is 216.