The lines (x-2)/2 = y/-2 = (z-7)/16 and (x+3)/4 = (y+2)/3 = (z+2)/1 intersect at the point P. If the

Q: Why do we use two different parameters, $\lambda$ and $\mu$, for the first two lines?

Great question! Each line is an independent entity in 3D space. A point on the first line is defined by its own parameter $\lambda$, and a point on the second line is defined by $\mu$. If we used the same parameter for both, we would be forcing the points to be at the same 'relative' position on both lines, which is not necessarily true. By using $\lambda$ and $\mu$, we allow the lines to intersect at any point $P$ that satisfies both equations independently.

Q: Can I use the cross product formula for the distance instead of the projection method?

Absolutely! The formula $l = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$ is perfectly valid and often taught in textbooks. The projection method I used is just another way to visualize the same geometry—it's essentially the same math, just viewed through the lens of a right-angled triangle. Use whichever method feels more intuitive to you, as long as you are careful with your vector calculations!

Q: What if the lines don't intersect? How would I know?

If the lines were skew (not intersecting and not parallel), the system of equations we set up in Step 3 would be inconsistent—meaning you would get a contradiction like $0 = 5$. In this problem, we are told they intersect at $P$, so we are guaranteed a consistent solution. Always check for consistency if you are unsure!

Question

The lines $\frac{x-2}{2} = \frac{y}{-2} = \frac{z-7}{16}$ and $\frac{x+3}{4} = \frac{y+2}{3} = \frac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\frac{x+1}{2} = \frac{y-1}{3} = \frac{z-1}{1}$ is $l$, then $14l^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Intersection of Paths

We begin by representing the first line, defined by the equation:

$$\frac{x-2}{2} = \frac{y}{-2} = \frac{z-7}{16}$$

Using a parameter $\lambda$, any point on this line can be expressed as $(2\lambda + 2, -2\lambda, 16\lambda + 7)$. Similarly, for the second line:

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

We introduce a parameter $\mu$ to represent any point on this line as $(4\mu - 3, 3\mu - 2, \mu - 2)$.

Since the lines intersect at point $P$, their coordinates must be identical at that specific location. Equating the $x$ and $y$ coordinates yields the following system:

$$2\lambda + 2 = 4\mu - 3$$
$$-2\lambda = 3\mu - 2$$

Solving this system by adding the equations, we find $2 = 7\mu - 5$, which simplifies to $7\mu = 7$, or $\mu = 1$. Substituting $\mu = 1$ back into the second equation gives $-2\lambda = 3(1) - 2$, resulting in $\lambda = -\frac{1}{2}$.

Plugging these parameters into our parametric forms, we determine the intersection point $P$ to be $(1, 1, -1)$.

### The Geometry of Distance

We now consider the third line, given by:

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

This line passes through a fixed point $A(-1, 1, 1)$ and has a direction vector $\vec{b} = 2\hat{i} + 3\hat{j} + \hat{k}$. We aim to find the perpendicular distance $l$ from point $P(1, 1, -1)$ to this line.

First, we define the vector $\vec{AP}$ connecting the fixed point $A$ to the intersection point $P$:

$$\vec{AP} = (1 - (-1))\hat{i} + (1 - 1)\hat{j} + (-1 - 1)\hat{k} = 2\hat{i} + 0\hat{j} - 2\hat{k}$$

The squared magnitude of this vector is $|\vec{AP}|^2 = 2^2 + 0^2 + (-2)^2 = 8$.

The perpendicular distance $l$ forms the height of a right-angled triangle where $\vec{AP}$ is the hypotenuse. The base of this triangle is the projection of $\vec{AP}$ onto the direction vector $\vec{b}$, calculated as $d = \frac{|\vec{AP} \cdot \vec{b}|}{|\vec{b}|}$.

The dot product is $\vec{AP} \cdot \vec{b} = (2)(2) + (0)(3) + (-2)(1) = 2$. The magnitude of the direction vector is $|\vec{b}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{14}$. Thus, the projection length is $d = \frac{2}{\sqrt{14}}$.

### The Final Calculation

Applying the Pythagorean theorem, we relate the distance $l$ to the hypotenuse and the projection:

$$l^2 = |\vec{AP}|^2 - d^2$$

Substituting our calculated values:

$$l^2 = 8 - \left(\frac{2}{\sqrt{14}}ight)^2 = 8 - \frac{4}{14} = \frac{112 - 4}{14} = \frac{108}{14}$$

The problem requires the value of $14l^2$. Multiplying our result by $14$:

$$14l^2 = 14 	imes \frac{108}{14} = 108$$

The final answer is **108**.

Enter Numerical Value:

Visualized Solution

Problem Overview

Parametric Form of Line 1

Parametric Form of Line 2

Equating Coordinates at $P$

Solving for Parameters

Coordinates of Point $P$

Analyzing the Third Line

Defining Vector $A P$

Magnitude of $A P$

Projection of $A P$ on Line 3

Calculating Distance Squared

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Intersection of Paths

The Geometry of Distance

The Final Calculation

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