If the lines (x-k)/1 = (y-2)/2 = (z-3)/3 and (x+1)/3 = (y+2)/2 = (z+3)/1 are co-planar, then the val

Q: Wait, why do we reject the idea of checking if they are parallel? Can't we just see if the direction vectors are proportional?

That is a great intuition, but checking for parallelism is only half the battle! If the lines are parallel, they are automatically coplanar. However, lines can also be coplanar if they intersect. By using the scalar triple product, we cover both cases—parallel lines and intersecting lines—in one elegant sweep. It is the universal condition for coplanarity.

Q: Why is the scalar triple product equal to zero? What does that actually mean?

Think of the scalar triple product as the volume of a parallelepiped formed by the three vectors. If the volume is zero, it means the parallelepiped has 'collapsed' into a flat shape. If the volume is zero, the vectors must lie in the same plane. It is a beautiful geometric interpretation of an algebraic condition!

Q: Does the order of the rows in the determinant matter?

Mathematically, swapping rows in a determinant only changes its sign. Since we are setting the entire determinant equal to zero, a change in sign does not affect the final result. You can arrange the vectors in any order, and you will arrive at the same value for $k$.

Question

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

Sigma Percentile · Accepted Answer

### The Geometry of 3D Space: A Journey into Coplanarity

Imagine you are standing in a vast, empty room with two thin, straight wires suspended in the air. In three-dimensional space, these wires could be parallel, intersecting, or 'skew'—meaning they never meet and are not parallel.

Today, we are interested in a special case: the condition where these two lines are **co-planar**. This means there exists a single, flat, infinite sheet of paper that can contain both lines simultaneously. It is a beautiful geometric constraint and a classic favorite of JEE examiners.

### The DNA of the Lines

To solve this, we first identify the 'DNA' of our lines. Every line in 3D space is defined by a point it passes through and a direction vector.

For the first line, $\frac{x-k}{1} = \frac{y-2}{2} = \frac{z-3}{3}$, we extract a point $A = (k, 2, 3)$ and a direction vector $\vec{b_1} = \hat{i} + 2\hat{j} + 3\hat{k}$.

For the second line, $\frac{x+1}{3} = \frac{y+2}{2} = \frac{z+3}{1}$, we find a point $B = (-1, -2, -3)$ and a direction vector $\vec{b_2} = 3\hat{i} + 2\hat{j} + \hat{k}$. We have now successfully extracted the essential information from both lines.

### The Bridge: The Connecting Vector $\vec{AB}$

To determine if these lines share a plane, we create a bridge between them. We define a vector $\vec{AB}$ that starts at point $B$ and ends at point $A$.

Calculating this via subtraction:
$$\vec{AB} = (k - (-1))\hat{i} + (2 - (-2))\hat{j} + (3 - (-3))\hat{k}$$

This simplifies to:
$$\vec{AB} = (k+1)\hat{i} + 4\hat{j} + 6\hat{k}$$

### The Scalar Triple Product: The Moment of Truth

The core principle is that if these two lines are coplanar, the two direction vectors $\vec{b_1}$ and $\vec{b_2}$ and the connecting vector $\vec{AB}$ must all lie in the same plane. Consequently, the volume of the parallelepiped they form must be zero.

Mathematically, this is expressed as the scalar triple product: $[\vec{AB} \quad \vec{b_1} \quad \vec{b_2}] = 0$. We represent this as a determinant:

$$\begin{vmatrix} k+1 & 4 & 6 \ 1 & 2 & 3 \ 3 & 2 & 1 \end{vmatrix} = 0$$

### The Determinant Dance

Now, we expand this determinant along the first row. Precision is paramount here.

The first term is $(k+1) 	imes (2(1) - 3(2)) = (k+1)(2 - 6) = -4(k+1)$.

The second term, applying the alternating sign, is $-4 	imes (1(1) - 3(3)) = -4(1 - 9) = -4(-8) = 32$.

The third term is $+6 	imes (1(2) - 2(3)) = 6(2 - 6) = 6(-4) = -24$.

Combining these, we get the equation:
$$-4(k+1) + 32 - 24 = 0$$

Simplifying further:
$$-4(k+1) + 8 = 0$$
$$-4k - 4 + 8 = 0$$
$$-4k + 4 = 0$$

Solving for $k$, we find:
**$k = 1$**

### Conclusion

The value of $k$ that forces these two lines into the same plane is **$1$**. This problem demonstrates how vector algebra simplifies complex geometric constraints.

Always remember: when you see 'coplanar lines' in a JEE problem, think of the scalar triple product. It is your most powerful weapon.

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

Enter Numerical Value:

Visualized Solution

Visualizing Coplanar Lines

Extracting Point and Direction of Line 1

Extracting Point and Direction of Line 2

Finding the Connecting Vector $A B$

The Condition for Coplanarity

Setting up the Determinant

Expanding the Determinant: First Term

Expanding the Determinant: Second Term

Expanding the Determinant: Third Term

Combining and Simplifying the Equation

Solving for $k$

Final Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of 3D Space

The DNA of the Lines

The Bridge

The Scalar Triple Product

The Determinant Dance

Conclusion

Similar Questions

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if

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If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{6}$ , then the sum of all possible values of $α$ is

If the shortest distance between the lines $\frac{x - λ}{- 2} = \frac{y - 2}{1} = \frac{z - 1}{1}$ and $\frac{x - 3}{1} = \frac{y - 1}{- 2} = \frac{z - 2}{1}$ is 1, then the sum of all possible values of $λ$ is :

Comprehension Passage

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The sum of all values of $α$ , for which the shortest distance between the lines $\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α}$ and $\frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $2$ , is

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is $\frac{1}{3}$ , then the sum of all possible values of $λ$ is :