If the lines r = (i - j + k) + λ(3j - k) and r = (αi - j) + μ(2i - 3k) are co-planar, then distance

Q: Wait, why do we use the scalar triple product to check for coplanarity? Can't we just use the shortest distance formula between two lines?

You absolutely could use the shortest distance formula! If the shortest distance between two lines is zero, they are coplanar. However, the scalar triple product is a much more elegant and direct tool when you have an unknown variable like $\alpha$ embedded in the coordinates. It turns a geometric condition into a simple algebraic determinant, which is much faster to solve.

Q: Does the order of the cross product matter when finding the normal vector $\vec{n}$?

Great question! The cross product $\vec{b_1} \times \vec{b_2}$ gives you a vector perpendicular to the plane. If you accidentally did $\vec{b_2} \times \vec{b_1}$, you would get the exact same vector but with all signs flipped (e.g., $-\vec{n}$ instead of $\vec{n}$). Since the equation of the plane $ax + by + cz + d = 0$ is equivalent to $-ax - by - cz - d = 0$, the final distance calculation remains identical. So, don't worry—the math is robust!

Q: Why do we use point $A$ to write the plane equation instead of point $B$?

It is purely a matter of convenience! A plane is defined by a normal vector and *any* point that lies on it. Since both lines lie on the plane, any point on line $L_1$ (like $A$) or any point on line $L_2$ (like $B$) will satisfy the plane's equation. Using $A$ is often simpler because it doesn't contain the variable $\alpha$, but using $B$ would yield the exact same result.

Question

If the lines $\vec{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda(3\hat{j} - \hat{k})$ and $\vec{r} = (\alpha\hat{i} - \hat{j}) + \mu(2\hat{i} - 3\hat{k})$ are co-planar, then distance of the plane containing these two lines from the point $(\alpha, 0, 0)$ is :

Options:

(A) $\frac{2}{9}$

(B) $\frac{2}{11}$

(C) $\frac{4}{11}$

(D) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are floating in a vast, three-dimensional void. You have two lines, $L_1$ and $L_2$, drifting through this space. They might seem like they are destined to never meet, but the problem whispers a secret: they are coplanar.

They lie on the same flat, infinite sheet. This is the starting point of our adventure. To solve this, we must first master the language of vectors.

We are given $L_1: \vec{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda(3\hat{j} - \hat{k})$ and $L_2: \vec{r} = (\alpha\hat{i} - \hat{j}) + \mu(2\hat{i} - 3\hat{k})$. Our first task is to identify the anchor points and the direction vectors.

For $L_1$, the point is $A(1, -1, 1)$ and the direction is $\vec{b_1} = (0, 3, -1)$. For $L_2$, the point is $B(\alpha, -1, 0)$ and the direction is $\vec{b_2} = (2, 0, -3)$.

### The Master Key: The Scalar Triple Product

Why does coplanarity matter? Because if two lines are coplanar, the vector connecting them, $\vec{AB}$, must lie in the same plane as their direction vectors, $\vec{b_1}$ and $\vec{b_2}$.

If you imagine these three vectors forming the edges of a parallelepiped, the volume of that shape must be zero. This is the physical intuition behind the scalar triple product: $[\vec{AB}, \vec{b_1}, \vec{b_2}] = 0$.

Let us calculate $\vec{AB} = B - A = (\alpha - 1, 0, -1)$. Now, we assemble our determinant:

$$\begin{vmatrix} \alpha - 1 & 0 & -1 \ 0 & 3 & -1 \ 2 & 0 & -3 \end{vmatrix} = 0$$

This determinant is the gatekeeper. Expanding along the first row, we get $(\alpha - 1)[3(-3) - 0(-1)] - 0 + (-1)[0(0) - 2(3)] = 0$.

Simplifying this, we find $(\alpha - 1)(-9) - 1(-6) = 0$, which leads us to $-9\alpha + 9 + 6 = 0$. Solving for $\alpha$, we get $-9\alpha + 15 = 0$, so $\alpha = \frac{15}{9} = \frac{5}{3}$. We have found our missing coordinate!

### Constructing the Plane

Now that we have $\alpha$, we need the equation of the plane itself. To define a plane, we need a point (we have $A(1, -1, 1)$) and a normal vector $\vec{n}$.

Since the plane contains both lines, $\vec{n}$ must be perpendicular to both $\vec{b_1}$ and $\vec{b_2}$. We find this using the cross product:

$$\vec{n} = \vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 3 & -1 \ 2 & 0 & -3 \end{vmatrix}$$

Expanding this, we get $\hat{i}(3(-3) - 0) - \hat{j}(0 - 2(-1)) + \hat{k}(0 - 3(2)) = -9\hat{i} - 2\hat{j} - 6\hat{k}$. For convenience, we can use $\vec{n} = (9, 2, 6)$.

The equation of the plane is $9(x - 1) + 2(y + 1) + 6(z - 1) = 0$, which simplifies to $9x + 2y + 6z - 13 = 0$.

### The Final Leap: Distance to the Point

Finally, we need the distance from the point $P(\alpha, 0, 0)$ to this plane. Since $\alpha = \frac{5}{3}$, our point is $P(\frac{5}{3}, 0, 0)$.

The distance formula is $d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$. Substituting our values:

$$d = \frac{|9(\frac{5}{3}) + 2(0) + 6(0) - 13|}{\sqrt{9^2 + 2^2 + 6^2}}$$

The numerator becomes $|15 - 13| = 2$. The denominator is $\sqrt{81 + 4 + 36} = \sqrt{121} = 11$.

Thus, the distance is **$\frac{2}{11}$**. We have traversed the 3D landscape and arrived at the solution.

Select Answer:

Visualized Solution

Analyzing the Given Lines

Condition for Coplanarity

Setting up the Determinant

Expanding the Determinant

Solving for $α$

Finding the Normal Vector $n$

Calculating the Normal Vector

Equation of the Plane

Simplifying the Plane Equation

Distance from a Point to a Plane

Applying the Distance Formula

Calculating the Final Distance

Final Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Master Key

Constructing the Plane

The Final Leap

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