Let the line passing through the points (-1,2,1) and parallel to the line (x-1)/2 = (y+1)/3 = z/4 in

Q: Wait, why do we use two different parameters $t$ and $s$ for the two lines? Can't we just use $t$ for both?

That is a fantastic question! If you use the same parameter $t$ for both lines, you are essentially forcing the lines to be at the same 'position' along their paths at the same time, which is not what we want. We just need to find the point where the two lines cross in 3D space. Since the lines are independent, the point $P$ will correspond to some value $t$ on $L_1$ and some value $s$ on $L_2$. They don't have to be the same value!

Q: What if the system of equations for $t$ and $s$ has no solution? Does that mean I made a mistake?

Not necessarily! If the system has no solution, it means the two lines are skew—they never intersect. However, in JEE problems of this type, the lines are usually designed to intersect. If you find no solution, double-check your arithmetic. If you are certain your math is correct, then the lines are indeed skew, and the distance between them would be a different, more complex problem involving the shortest distance between skew lines.

Q: Why do we only equate the $x$ and $y$ coordinates to solve for $t$ and $s$? What about $z$?

We use the $x$ and $y$ equations to solve for the two unknowns ($t$ and $s$) because two equations are sufficient to solve for two variables. The $z$ coordinate is then used as a 'check' to ensure the intersection point is consistent across all three dimensions. If the $t$ and $s$ values you found don't satisfy the $z$ equation, it means the lines do not intersect at a single point.

Question

Let the line passing through the points $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ intersect the line $\frac{x+2}{3}=rac{y-3}{2}=rac{z-4}{1}$ at the point P. Then the distance of P from the point $Q(4,-5,1)$ is

Options:

(A) 5

(B) $5\sqrt{5}$

(C) $5\sqrt{6}$

(D) 10

Sigma Percentile · Accepted Answer

### The Geometry of Collision: Finding the Intersection in 3D Space

Imagine you are standing in a vast, three-dimensional void. You see two lines stretching out into infinity, like two contrails of airplanes crossing in the sky.

One line, $L_1$, passes through the point $A(-1, 2, 1)$ and is parallel to the line defined by:

$$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z}{4}$$

The other line, $L_2$, is given by the equation:

$$\frac{x+2}{3} = \frac{y-3}{2} = \frac{z-4}{1}$$

Your mission is to find the point $P$ where these two lines intersect and then calculate the distance from this point to a third point $Q(4, -5, 1)$.

### Phase 1: Defining the Paths

To understand a line in 3D, we need two things: a starting point and a direction. For $L_1$, we are given the point $A(-1, 2, 1)$.

The problem states $L_1$ is parallel to the line with direction ratios $(2, 3, 4)$. Thus, the direction vector of $L_1$ is $\vec{d_1} = (2, 3, 4)$.

We can write the symmetric equation of $L_1$ as:

$$\frac{x+1}{2} = \frac{y-2}{3} = \frac{z-1}{4}$$

By setting this expression equal to a parameter $t$, we describe any point on $L_1$ as $(2t - 1, 3t + 2, 4t + 1)$. This represents our general point $P$ on $L_1$.

### Phase 2: The Collision

Now, we turn our attention to $L_2$. We introduce a different parameter, $s$, because the lines are independent entities.

Setting the equation of $L_2$ equal to $s$, we find the general point on $L_2$ to be $(3s - 2, 2s + 3, s + 4)$. For the lines to intersect at point $P$, the coordinates must be identical at that specific location.

This gives us a system of equations:

$$2t - 1 = 3s - 2 \Rightarrow 2t - 3s = -1$$

$$3t + 2 = 2s + 3 \Rightarrow 3t - 2s = 1$$

Using the method of elimination, we multiply the first equation by $2$ and the second by $3$ to align the coefficients of $s$. This yields $4t - 6s = -2$ and $9t - 6s = 3$.

Subtracting these equations, we find $5t = 5$, which gives us $t = 1$. Substituting $t = 1$ back into our second equation, $3(1) - 2s = 1$, we find $s = 1$.

### Phase 3: The Final Destination

With $t = 1$, we find the exact coordinates of the intersection point $P$. Plugging $t = 1$ into our expression for $L_1$, we get $x = 2(1) - 1 = 1$, $y = 3(1) + 2 = 5$, and $z = 4(1) + 1 = 5$.

So, the intersection point is $P(1, 5, 5)$. Now, we calculate the distance between $P(1, 5, 5)$ and $Q(4, -5, 1)$ using the 3D distance formula:

$$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

Substituting our coordinates, we get:

$$PQ = \sqrt{(4 - 1)^2 + (-5 - 5)^2 + (1 - 5)^2}$$

This simplifies to:

$$PQ = \sqrt{3^2 + (-10)^2 + (-4)^2} = \sqrt{9 + 100 + 16} = \sqrt{125}$$

Factoring this, we get $\sqrt{25 	imes 5} = 5\sqrt{5}$. The final distance is **$5\sqrt{5}$**.

Select Answer:

Visualized Solution

The Geometric Setup

Direction of Line $L_{1}$

Equation of Line $L_{1}$

Equation of Line $L_{2}$

Equating Coordinates at $P$

Solving for Parameters

Coordinates of Point $P$

The Final Objective

Applying Distance Formula

Calculating the Distance

Final Answer

The Sigma Insight: Equation of a Line in Space

The Geometry of Collision

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

The Geometric Setup

Direction of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Equating Coordinates at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P