If the lines (x-1)/2 = (y+1)/3 = (z-1)/4 and (x-3)/1 = (y-k)/2 = z/1 intersect, then the value of k

Q: Wait, why can't we just solve for x, y, and z by setting the lines equal to each other?

You absolutely can! You could set both lines equal to parameters $\lambda$ and $\mu$, solve for the variables, and see if they yield a consistent point. However, the scalar triple product method is a much more elegant and faster way to find the unknown variable $k$ without needing to find the exact point of intersection.

Q: What if the lines are parallel? Does the coplanarity condition still work?

Great question! If the lines are parallel, their direction vectors are proportional. In that case, the determinant would also be zero. However, in this problem, the direction vectors $(2, 3, 4)$ and $(1, 2, 1)$ are clearly not proportional, so we know they aren't parallel. The zero determinant confirms they intersect.

Q: Why does the scalar triple product have to be zero for intersection?

The scalar triple product represents the volume of a parallelepiped formed by three vectors. If the lines intersect, the vector connecting the two lines and the two direction vectors all lie in the same plane. A flat 2D shape has zero volume in 3D space, hence the determinant must be zero.

Question

If the lines $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}$ and $\frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1}$ intersect, then the value of $k$ is

Options:

(A) $3/2$

(B) $9/2$

(C) $-2/9$

(D) $-3/2$

Sigma Percentile · Accepted Answer

### The Geometry of Intersection

Imagine standing in a vast, empty room with two laser pointers casting beams of light. These beams represent lines in 3D space. Most randomly oriented lines are skew, meaning they pass each other without ever touching.

For two lines to intersect in 3D, they must be **coplanar**. They must lie on the same invisible, flat sheet of paper. If they are not on the same plane, they can never touch.

### The Mathematical Engine

To force these lines to intersect, we use the **Scalar Triple Product** as a tool to detect coplanarity. Consider the two direction vectors $\vec{b}_1 = (2, 3, 4)$ and $\vec{b}_2 = (1, 2, 1)$, along with a bridge vector $\vec{AB}$ connecting a point on the first line to a point on the second.

If these three vectors lie in the same plane, they cannot form a 3D volume. The volume of the parallelepiped they define is given by the determinant of the matrix formed by these vectors. If that volume is zero, the lines are coplanar and, therefore, they intersect.

### Extracting the DNA of the Lines

First, we analyze the given lines. Line 1 is defined by:

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

From this, we extract a point $A(1, -1, 1)$ and a direction vector $\vec{b}_1 = (2, 3, 4)$.

Line 2 is defined by:

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

Here, we identify a point $B(3, k, 0)$ and a direction vector $\vec{b}_2 = (1, 2, 1)$. The variable $k$ is the parameter we must determine to ensure intersection.

### Building the Bridge

We define the bridge vector $\vec{AB}$ connecting point $A$ to point $B$. We calculate this by subtracting the coordinates of $A$ from $B$:

$$\vec{AB} = (3-1, k-(-1), 0-1) = (2, k+1, -1)$$

Now, we construct the determinant condition for coplanarity:

$$\begin{vmatrix} 2 & k+1 & -1 \ 2 & 3 & 4 \ 1 & 2 & 1 \end{vmatrix} = 0$$

### The Final Calculation

Expanding this determinant along the first row, we obtain:

$$2 \cdot \begin{vmatrix} 3 & 4 \ 2 & 1 \end{vmatrix} - (k+1) \cdot \begin{vmatrix} 2 & 4 \ 1 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & 3 \ 1 & 2 \end{vmatrix} = 0$$

Evaluating the minors, we get:

$$2(3 - 8) - (k+1)(2 - 4) - 1(4 - 3) = 0$$

This simplifies to:

$$2(-5) - (k+1)(-2) - 1(1) = 0$$

$$-10 + 2(k+1) - 1 = 0$$

$$-11 + 2k + 2 = 0$$

$$2k - 9 = 0$$

Solving for $k$, we find the final result:

**$k = \frac{9}{2}$**

By enforcing the condition of coplanarity, we have determined the exact value of $k$ that forces these two laser beams to meet in 3D space.

Animated Solution for Mathematics - Three Dimensional Geometry: If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then the value of $k$ is

Select Answer:

Visualized Solution

Visualizing Intersecting Lines

The Coplanarity Condition

The Mathematical Tool

Extracting Line 1 Parameters

Extracting Line 2 Parameters

The Connecting Vector

Setting up the Determinant

Expanding the Determinant

Evaluating the Minors

Simplifying the Equation

Solving for $k$

Final Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of Intersection

The Mathematical Engine

Extracting the DNA of the Lines

Building the Bridge

The Final Calculation

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