JEE Mathematics 2025 Question with Solution

Given:

• $L_1 : \frac{x - 1}{3} = \frac{y - 1}{-1} = \frac{z + 1}{0}$
• $L_2 : \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{\alpha}$
• Point $A(1,1,-1)$

Find: $26\alpha (PB)^2$, where $B$ is the intersection point of the two lines and $P$ is the foot of the perpendicular from $A$ on $L_2$.

Write the lines in parametric form:

$$x = 1 + 3s, \quad y = 1 - s, \quad z = -1$$

for $L_1$, and

$$x = 2 + 2t, \quad y = 0, \quad z = -4 + \alpha t$$

for $L_2$.

Since the lines intersect at $B$, use the common point conditions. From $y=0$ on $L_2$ and $y=1-s$ on $L_1$,

$$1-s=0 \Rightarrow s=1$$

Hence on $L_1$,

$$B=(1+3,0,-1)=(4,0,-1)$$

Now on $L_2$, the point $B$ must satisfy

$$2+2t=4 \Rightarrow t=1$$

and

$$-4+\alpha t=-1$$

So,

$$-4+\alpha=-1 \Rightarrow \alpha=3$$

Therefore, direction vector of $L_2$ is

$$\vec d=(2,0,3)$$

and a point on $L_2$ is $D_0=(2,0,-4)$.

Let the foot of the perpendicular be

$$P=D_0+t\vec d$$

Since $AP \perp L_2$,

$$(A-P)\cdot \vec d=0$$

Using the projection formula,

$$t=\frac{(A-D_0)\cdot \vec d}{\vec d\cdot \vec d}$$

Now,

$$A-D_0=(1-2,1-0,-1+4)=(-1,1,3)$$

So,

$$(A-D_0)\cdot \vec d=(-1)(2)+(1)(0)+(3)(3)=7$$

and

$$\vec d\cdot \vec d=2^2+0^2+3^2=13$$

Thus,

$$t=\frac{7}{13}$$

Hence,

$$P=(2,0,-4)+\frac{7}{13}(2,0,3)=\left(\frac{40}{13},0,-\frac{31}{13}\right)$$

Now,

$$\overrightarrow{PB}=B-P=\left(4-\frac{40}{13},0,-1+\frac{31}{13}\right)=\left(\frac{12}{13},0,\frac{18}{13}\right)$$

Therefore,

$$(PB)^2=\left(\frac{12}{13}\right)^2+\left(\frac{18}{13}\right)^2=\frac{144+324}{169}=\frac{468}{169}=\frac{36}{13}$$

Finally,

$$26\alpha (PB)^2=26\cdot 3 \cdot \frac{36}{13}=216$$

Therefore, the value of $26\alpha (PB)^2$ is $216$.

Given: The two lines intersect, and $P$ is the foot of the perpendicular from $A(1,1,-1)$ to $L_2$.

Find: $26\alpha (PB)^2$.

Interpret the zero denominators carefully. From $L_1$,

$$\frac{z+1}{0}$$

means

$$z=-1$$

so a point on $L_1$ is $(1,1,-1)$ and its direction vector is

$$(3,-1,0)$$

From $L_2$,

$$\frac{y}{0}$$

means

$$y=0$$

so a point on $L_2$ is $(2,0,-4)$ and its direction vector is

$$(2,0,\alpha)$$

Parametrize $L_1$ as

$$x=1+3s, \quad y=1-s, \quad z=-1$$

At the intersection point, $y=0$, so

$$1-s=0 \Rightarrow s=1$$

Hence,

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

Now parametrize $L_2$ as

$$x=2+2t, \quad y=0, \quad z=-4+\alpha t$$

Since $B$ lies on $L_2$,

$$2+2t=4 \Rightarrow t=1$$

Then,

$$-4+\alpha(1)=-1 \Rightarrow \alpha=3$$

So $L_2$ becomes the line through $(2,0,-4)$ in the direction $\vec d=(2,0,3)$.

Let

$$P=(2+2t,0,-4+3t)$$

Because $P$ is the foot of the perpendicular from $A$, vector $AP$ is perpendicular to $\vec d$. Thus,

$$\big((2+2t)-1,0-1,(-4+3t)+1\big)\cdot (2,0,3)=0$$

That is,

$$(1+2t,-1,-3+3t)\cdot (2,0,3)=0$$

So,

$$2(1+2t)+3(-3+3t)=0$$ $$2+4t-9+9t=0$$ $$13t=7$$ $$t=\frac{7}{13}$$

Hence,

$$P=\left(2+\frac{14}{13},0,-4+\frac{21}{13}\right)=\left(\frac{40}{13},0,-\frac{31}{13}\right)$$

Now,

$$\overrightarrow{PB}=\left(4-\frac{40}{13},0,-1+\frac{31}{13}\right)=\left(\frac{12}{13},0,\frac{18}{13}\right)$$

Therefore,

$$(PB)^2=\left(\frac{12}{13}\right)^2+\left(\frac{18}{13}\right)^2=\frac{468}{169}=\frac{36}{13}$$

Thus,

$$26\alpha (PB)^2=26\cdot 3 \cdot \frac{36}{13}=216$$

Therefore, the required value is $216$.