Let O be the origin, and M and N be the points on the lines (x-5)/4 = (y-4)/1 = (z-5)/3 and (x+8)/12

Q: Wait, why do we need to find the points $M$ and $N$ individually? Can't we just use the shortest distance formula?

That is a brilliant question! The shortest distance formula gives you the *length* of the segment, but the question asks for the dot product $\vec{OM} \cdot \vec{ON}$. To compute this, you absolutely need the coordinates of $M$ and $N$. Think of the distance formula as a shortcut that skips the very information you need here. We must find the points to solve the problem.

Q: Why did the $j$-component of the cross product become zero? Did I make a mistake in my determinant calculation?

Not at all! A zero component is perfectly valid and often appears in these problems. It simply means the common perpendicular vector lies in the $xz$-plane. It actually makes your algebra easier, not harder. Trust your calculation; the math is guiding you toward a simpler path.

Q: How do I know if the lines are skew or intersecting without checking the distance first?

In JEE problems, if you are asked to find the shortest distance, they are almost certainly skew. However, if you were to solve the system of equations for $\lambda$ and $\mu$ and found no solution, that would imply the lines are parallel. If you found a solution where the distance was zero, they would intersect. The algebra will always reveal the truth if you follow the steps rigorously.

Question

Let $O$ be the origin, and $M$ and $N$ be the points on the lines $\frac{x-5}{4} = \frac{y-4}{1} = \frac{z-5}{3}$ and $\frac{x+8}{12} = \frac{y+2}{5} = \frac{z+11}{9}$ respectively such that $MN$ is the shortest distance between the given lines. Then $\vec{OM} \cdot \vec{ON}$ is equal to _____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To navigate 3D space, we need a coordinate system. We are given two lines, $L_1$ and $L_2$. To find any point on these lines, we use the power of parameters.

By setting the equations of the lines equal to a scalar—let us call them $\lambda$ and $\mu$—we give every point on the line an 'address'.

For line $L_1$, defined by 
$$\frac{x-5}{4} = \frac{y-4}{1} = \frac{z-5}{3} = \lambda$$ 
we can express any point $M$ as $(4\lambda + 5, \lambda + 4, 3\lambda + 5)$.

Similarly, for line $L_2$, defined by 
$$\frac{x+8}{12} = \frac{y+2}{5} = \frac{z+11}{9} = \mu$$ 
any point $N$ is $(12\mu - 8, 5\mu - 2, 9\mu - 11)$. By doing this, we have transformed a static line into a dynamic variable.

### The Bridge of Perpendicularity

Now, we define the vector $\vec{MN}$, which connects our two points. This vector is simply $\vec{N} - \vec{M}$. When we perform the subtraction, we get:

$$\vec{MN} = (12\mu - 4\lambda - 13)\hat{i} + (5\mu - \lambda - 6)\hat{j} + (9\mu - 3\lambda - 16)\hat{k}$$

For $\vec{MN}$ to be the **shortest distance**, it must be the common perpendicular. It must be perpendicular to the direction of $L_1$ ($\vec{d_1} = 4\hat{i} + 1\hat{j} + 3\hat{k}$) and the direction of $L_2$ ($\vec{d_2} = 12\hat{i} + 5\hat{j} + 9\hat{k}$).

We calculate the cross product $\vec{d_1} 	imes \vec{d_2}$ to find the direction of this bridge:

$$\vec{d_1} 	imes \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 1 & 3 \ 12 & 5 & 9 \end{vmatrix} = -6\hat{i} + 0\hat{j} + 8\hat{k}$$

This vector, $(-6, 0, 8)$, is the direction of our shortest distance. It is the compass pointing us toward the solution.

### Solving the System

Since $\vec{MN}$ must be parallel to this cross product, the components of $\vec{MN}$ must be proportional to $(-6, 0, 8)$. Because the cross product has a zero in the $j$-direction, the $y$-component of $\vec{MN}$ must also be zero:

$$5\mu - \lambda - 6 = 0 \implies \lambda = 5\mu - 6$$

Substituting this relationship into the other components, we solve the system to find $\mu = 1$ and $\lambda = -1$. We have now identified the exact locations of $M$ and $N$ in space.

### Final Calculation

With $\lambda = -1$, point $M$ becomes $(1, 3, 2)$. With $\mu = 1$, point $N$ becomes $(4, 3, -2)$.

These are the coordinates of our points relative to the origin $O$. The vectors $\vec{OM}$ and $\vec{ON}$ are simply the position vectors of these points. Finally, we compute the dot product:

$$\vec{OM} \cdot \vec{ON} = (1)(4) + (3)(3) + (2)(-2) = 4 + 9 - 4 = 9$$

The final result is **9**. You have successfully navigated the skew lines, found the common perpendicular, and arrived at the destination.

Enter Numerical Value:

Visualized Solution

Visualizing the Skew Lines

Parametric Coordinates of Point $M$

Parametric Coordinates of Point $N$

Defining Vector $M N$

The Common Perpendicular Condition

Calculating the Cross Product

Setting up the Parallelism

Solving for $λ$ and $μ$

Finding Coordinates of $M$

Finding Coordinates of $N$

Calculating the Dot Product

Summary and Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Bridge of Perpendicularity

Solving the System

Final Calculation

Similar Questions

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to ______.

If the shortest distance between the lines $r_{1} = α \hat{i} + 2 \hat{j} + 2 \hat{k} + λ (\hat{i} - 2 \hat{j} + 2 \hat{k}), λ \in R, α > 0$ and $r_{2} = - 4 \hat{i} - \hat{k} + μ (3 \hat{i} - 2 \hat{j} - 2 \hat{k}), μ \in R$ is 9, then $α$ is equal to ___

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is

The shortest distance between the lines $\frac{x - 3}{4} = \frac{y + 7}{- 11} = \frac{z - 1}{5}$ and $\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z + 2}{1}$ is:

Shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ is

Enter Numerical Value:

Visualized Solution

Visualizing the Skew Lines

Parametric Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Parametric Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Defining Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MN