Let the line L pass through (1, 1, 1) and intersect the lines (x-1)/2 = (y+1)/3 = (z-1)/4 and (x-3)/

Q: Wait, why can't we just use the parametric form of the line $L$?

You *could*, but it's algebraically tedious. The plane intersection method is much faster and less prone to calculation errors. In JEE, speed and accuracy are your best friends. By finding the two planes, we bypass the need to solve for parameters entirely.

Q: How do I know which two vectors to cross to get the normal of the plane?

You need two vectors lying in the plane. One is the direction vector of the line, and the other is the vector connecting the fixed point $P$ to any point on the line. Their cross product gives the normal vector perpendicular to both, which is exactly what you need for the plane equation.

Q: What if the lines were parallel?

If the lines were parallel, the planes might be the same or undefined, but in this specific problem, they are skew/intersecting in a way that creates distinct planes. Always check the direction vectors first to ensure they aren't parallel!

Question

Let the line $L$ pass through $(1, 1, 1)$ and intersect the lines $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}$ and $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z}{1}$. Then, which of the following points lies on the line $L$ ?

Options:

(A) $(4, 22, 7)$

(B) $(5, 4, 3)$

(C) $(10, -29, -50)$

(D) $(7, 15, 13)$

Sigma Percentile · Accepted Answer

### The Geometry of the Invisible Line

Imagine standing in a vast, three-dimensional space. You are anchored at a specific point $P(1, 1, 1)$.

Floating around you are two distinct lines, $L_1$ and $L_2$. Your mission is to construct a new line, $L$, that passes through your anchor point $P$ and pierces through both $L_1$ and $L_2$.

At first glance, this feels like trying to thread a needle in the dark. But here is the secret: we don't need to hunt for the line directly. We can use the power of planes to trap it.

### The Plane Trap

Think about the relationship between a line and a plane. If our target line $L$ must pass through $P$ and intersect $L_1$, then $L$ must lie entirely within the flat surface—the plane—that contains both $P$ and $L_1$.

Let's call this plane $\pi_1$. By the exact same logic, $L$ must also lie within the plane $\pi_2$ that contains $P$ and $L_2$.

If $L$ is trapped in $\pi_1$ and also trapped in $\pi_2$, then $L$ must be the line where these two planes intersect. This is the geometric epiphany that turns a nightmare problem into a simple, elegant calculation.

### Constructing the Planes

To define $\pi_1$, we need a point and a normal vector. We have point $P(1, 1, 1)$. From the equation of $L_1$, we can extract a point $A(1, -1, 1)$ and a direction vector $\vec{d_1} = (2, 3, 4)$.

The vector connecting our anchor $P$ to point $A$ is $\vec{PA} = A - P = (0, -2, 0)$. To find the normal vector $\vec{n_1}$, we take the cross product of $\vec{PA}$ and $\vec{d_1}$:

$$\vec{n_1} = \vec{PA} 	imes \vec{d_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -2 & 0 \ 2 & 3 & 4 \end{vmatrix} = -8\hat{i} + 4\hat{k}$$

We can simplify this normal vector to $\vec{n_1} = (2, 0, -1)$. Using the point-normal form of a plane, $\vec{n_1} \cdot (\vec{r} - \vec{P}) = 0$, we get $2(x-1) + 0(y-1) - 1(z-1) = 0$.

This simplifies beautifully to $\pi_1: 2x - z - 1 = 0$.

Now, we repeat this for $\pi_2$. $L_2$ gives us a point $B(3, 4, 0)$ and direction $\vec{d_2} = (1, 2, 1)$. The vector $\vec{PB} = B - P = (2, 3, -1)$.

The normal $\vec{n_2}$ is the cross product of $\vec{PB}$ and $\vec{d_2}$:

$$\vec{n_2} = \vec{PB} 	imes \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -1 \ 1 & 2 & 1 \end{vmatrix} = (3 - (-2))\hat{i} - (2 - (-1))\hat{j} + (4 - 3)\hat{k} = (5, -3, 1)$$

Using point $P$ again, the equation for $\pi_2$ becomes $5(x-1) - 3(y-1) + 1(z-1) = 0$. This simplifies to $\pi_2: 5x - 3y + z - 3 = 0$.

### The Final Verification

We have successfully trapped our line $L$ as the intersection of $\pi_1$ and $\pi_2$. Any point on $L$ must satisfy both $2x - z - 1 = 0$ and $5x - 3y + z - 3 = 0$.

Instead of solving for the line's parametric equation, we simply test our options. Plugging $(7, 15, 13)$ into $\pi_1$: 
$$2(7) - 13 - 1 = 14 - 14 = 0$$

It works! Plugging it into $\pi_2$: 
$$5(7) - 3(15) + 13 - 3 = 35 - 45 + 13 - 3 = 0$$

It works again! Geometry isn't just about formulas; it's about seeing the hidden structure of the world. By visualizing the planes, we didn't just solve a problem—we **mastered the space**.

Animated Solution for Mathematics - Three Dimensional Geometry: Let the line $L$ pass through $(1, 1, 1)$ and intersect the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1}$ . Then, which of the following points lies on the line $L$ ?

Select Answer:

Visualized Solution

Visualizing the Geometry

The Concept of Intersecting Planes

Defining Plane $π_{1}$

Normal Vector of $π_{1}$

Equation of Plane $π_{1}$

Defining Plane $π_{2}$

Normal Vector of $π_{2}$

Equation of Plane $π_{2}$

The Line of Intersection

Verifying the Options

Final Conclusion

The Sigma Insight: Equation of a Line in Space

The Geometry of the Invisible Line

The Plane Trap

Constructing the Planes

The Final Verification

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