If the shortest distance between the lines r1 = αi + 2j + 2k + λ(i - 2j + 2k), λ ∈ R, α 0 and r2 =

Q: Wait, why can't we just find the intersection point of these two lines?

That is a fantastic question! The reason we cannot find an intersection point is that these lines are skew. In three-dimensional space, skew lines are lines that are neither parallel nor intersecting. They are like two ships passing in the night at different altitudes; they will never meet. That is precisely why we need the shortest distance formula—to measure the gap between them.

Q: Why is there an absolute value in the shortest distance formula?

The absolute value is crucial because distance must always be a non-negative quantity. The scalar triple product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$ can result in a negative value depending on the order in which you subtract the vectors or calculate the cross product. Since distance is a physical length, we take the magnitude to ensure our result is positive.

Q: Why do we reject the negative root $\alpha = -21$?

Great catch! In many algebraic problems, we get multiple solutions, but we must always look back at the problem statement. The question explicitly states the constraint $\alpha > 0$. This is a physical or geometric constraint that limits our valid solutions. Always check your constraints at the end of a problem!

Question

If the shortest distance between the lines $\vec{r}_1 = \alpha \hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k}), \lambda \in \mathbb{R}, \alpha > 0$ and $\vec{r}_2 = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k}), \mu \in \mathbb{R}$ is 9, then $\alpha$ is equal to ___

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional room. You have two lines, $L_1$ and $L_2$, floating in space. They are not parallel, and they do not intersect.

These are what we call **skew lines**. They are like two paths that never cross, separated by a specific, minimum gap.

Our mission today is to find the value of the parameter $\alpha$ that dictates the position of one of these lines, given that the shortest distance between them is exactly $9$ units. This is not just algebra; it is the art of navigating 3D space.

### The Toolset: The Shortest Distance Formula

To bridge the gap between these lines, we need a powerful tool. The shortest distance $d$ between two lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by the formula:

$$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|}$$

Think of the numerator as the volume of a parallelepiped formed by the vectors, and the denominator as the area of its base. The ratio gives us the perpendicular height—the shortest distance.

### Extracting the Ingredients

First, we must carefully extract our vectors from the given equations.

For $L_1: \vec{r}_1 = (\alpha \hat{i} + 2\hat{j} + 2\hat{k}) + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})$, we identify:
$\vec{a}_1 = \alpha \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b}_1 = \hat{i} - 2\hat{j} + 2\hat{k}$.

For $L_2: \vec{r}_2 = (-4\hat{i} - \hat{k}) + \mu(3\hat{i} - 2\hat{j} - 2\hat{k})$, we have:
$\vec{a}_2 = -4\hat{i} + 0\hat{j} - \hat{k}$ and $\vec{b}_2 = 3\hat{i} - 2\hat{j} - 2\hat{k}$.

Now, we find the difference vector:
$$\vec{a}_2 - \vec{a}_1 = (-4 - \alpha)\hat{i} - 2\hat{j} - 3\hat{k}$$

### The Common Normal

Next, we find the common perpendicular vector using the cross product $\vec{b}_1 	imes \vec{b}_2$. We set up the determinant:

$$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 2 \ 3 & -2 & -2 \end{vmatrix}$$

Expanding this, we get $\hat{i}(4 - (-4)) - \hat{j}(-2 - 6) + \hat{k}(-2 - (-6))$, which simplifies to $8\hat{i} + 8\hat{j} + 4\hat{k}$.

The magnitude of this vector is:
$$|\vec{b}_1 	imes \vec{b}_2| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$$

### The Final Calculation

Now, we compute the dot product for the numerator:
$$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2) = (-4 - \alpha)(8) + (-2)(8) + (-3)(4) = -32 - 8\alpha - 16 - 12 = -8\alpha - 60$$

Substituting everything into our distance formula, we get:
$$9 = \frac{|-8\alpha - 60|}{12}$$

This leads to two cases: $2\alpha + 15 = 27$ or $2\alpha + 15 = -27$. Solving these, we get $\alpha = 6$ or $\alpha = -21$.

Given the constraint $\alpha > 0$, we reject the negative value. Thus, **$\alpha = 6$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Skew Lines

The Shortest Distance Formula

Extracting Vectors for Line 1

Extracting Vectors for Line 2

Calculating $a_{2} - a_{1}$

Setting up the Cross Product

Computing the Cross Product

Magnitude of the Cross Product

The Dot Product (Numerator)

Applying the Shortest Distance Formula

Simplifying the Equation

Solving for $α$

Final Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Toolset

Extracting the Ingredients

The Common Normal

The Final Calculation

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