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 Options: (A) $k = 3$ or $-2$ (B) $k = 0$ or $-1$ (C) $k = 1$ or $-1$ (D) $k = 0$ or $-3$

### Analyzing the Setup Imagine you are standing in a vast, empty 3D room. Two straight lines are floating in front of you, suspended in mid-air. We want to determine if these lines are coplanar—that is, if they can be contained within a single, flat sheet of paper. If the lines are skew, they exist in different "layers" of space and can never share a plane. To force these lines onto the same plane, we utilize the geometric property that the volume of the parallelepiped formed by their connecting vector and their direction vectors must be zero. ### The Mathematical Bridge We identify a point on the first line, $P_1(2, 3, 4)$, and a point on the second line, $P_2(1, 4, 5)$. We define the bridge vector connecting these points as $\vec{P_1P_2} = (1-2, 4-3, 5-4) = (-1, 1, 1)$. The direction vectors of the two lines are given as $\vec{d_1} = (1, 1, -k)$ and $\vec{d_2} = (k, 2, 1)$. For the lines to be coplanar, the scalar triple product of these three vectors must vanish: $$[\vec{P_1P_2} \quad \vec{d_1} \quad \vec{d_2}] = 0$$ ### The Determinant Dance We assemble the components into a determinant and set it to zero: $$\begin{vmatrix} -1 & 1 & 1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix} = 0$$ Expanding this determinant along the first row, we obtain: $$-1 \cdot \begin{vmatrix} 1 & -k \\ 2 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & -k \\ k & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 \\ k & 2 \end{vmatrix} = 0$$ Calculating each minor individually: 1. The first minor is $(1 - (-2k)) = (1 + 2k)$. 2. The second minor is $(1 - (-k^2)) = (1 + k^2)$. 3. The third minor is $(2 - k)$. ### The Collapse of Complexity Substituting these values back into our equation, we get: $$-1(1 + 2k) - 1(1 + k^2) + 1(2 - k) = 0$$ Expanding the terms yields: $$-1 - 2k - 1 - k^2 + 2 - k = 0$$ The constant terms $-1 - 1 + 2$ cancel out perfectly. We are left with the simplified quadratic equation: $$-k^2 - 3k = 0$$ Multiplying by $-1$ and factoring, we find: $$k(k + 3) = 0$$ This yields the final solutions: **$k = 0$** or **$k = -3$**. These values represent the specific conditions under which the two lines are coplanar.

JEE Main 2003

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: 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

Q: How do I know which point is $(x_1, y_1, z_1)$ and which is $(x_2, y_2, z_2)$? Does the order matter?

It does not matter! If you swap the points, the vector $\vec{P_1P_2}$ becomes $\vec{P_2P_1}$, which is just the negative of the original vector. Since the determinant is linear with respect to its rows, swapping the points only multiplies the entire determinant by $-1$. Since we are setting the determinant to zero, multiplying by $-1$ doesn't change the final values of $k$ at all. You are safe either way!

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the two lines as floating in 3D space. For them to be coplanar, they must either intersect or be parallel. This means they must lie on a single flat sheet of paper. If they are skew, they will never meet and cannot share a plane. The geometric condition for this is that the volume of the parallelepiped formed by the vector connecting a point on each line and the two direction vectors must be zero.

Hint 2

Hint 2 (The Arsenal): Recall the concept of the Scalar Triple Product. If a line passes through point

P_{1}

with direction vector

d_{1}

and another line passes through

P_{2}

with direction vector

d_{2}

, the condition for coplanarity is that the vectors

P_{1} P_{2}

d_{1}

, and

d_{2}

are linearly dependent. This is mathematically expressed as the determinant of the matrix formed by these three vectors being equal to zero.

Hint 3

Hint 3 (The Execution): Set up the determinant using the coordinates of the points and the components of the direction vectors:

x_{2} - x_{1} a_{1} a_{2} y_{2} - y_{1} b_{1} b_{2} z_{2} - z_{1} c_{1} c_{2} = 0

. Substitute the values from your lines into this determinant and solve the resulting equation for

k

Visualized Solution

Visualizing Coplanar Lines

Two lines in 3D space are coplanar if they lie entirely in the same plane.
This means they either intersect at a point or are perfectly parallel.
For skew lines, they never meet and don't share a plane. Here, we force them to share one!

The Condition for Coplanarity

Let the lines pass through points $P_{1}$ and $P_{2}$ , with direction vectors $d_{1}$ and $d_{2}$ .
The vector connecting the points is $P_{1} P_{2}$ .
For coplanarity, $P_{1} P_{2}$ , $d_{1}$ , and $d_{2}$ must lie in the same plane.
Mathematically, their scalar triple product must be zero: $[P_{1} P_{2} d_{1} d_{2}] = 0$ .

Parameters of Line 1

Line 1: $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$
Comparing with standard form $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}$ :
Point $P_{1} = (2, 3, 4)$
Direction $d_{1} = (1, 1, - k)$

Parameters of Line 2

Line 2: $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$
Point $P_{2} = (1, 4, 5)$
Direction $d_{2} = (k, 2, 1)$

Vector $P_{1} P_{2}$

We need the vector connecting $P_{1} (2, 3, 4)$ to $P_{2} (1, 4, 5)$ .
$P_{1} P_{2} = (x_{2} - x_{1}) \hat{i} + (y_{2} - y_{1}) \hat{j} + (z_{2} - z_{1}) \hat{k}$
$P_{1} P_{2} = (1 - 2) \hat{i} + (4 - 3) \hat{j} + (5 - 4) \hat{k}$
$P_{1} P_{2} = (- 1, 1, 1)$

Setting up the Determinant

The scalar triple product $[P_{1} P_{2} d_{1} d_{2}] = 0$ is computed using a determinant.
$x_{2} - x_{1} a_{1} a_{2} y_{2} - y_{1} b_{1} b_{2} z_{2} - z_{1} c_{1} c_{2} = 0$
Substituting our vectors:
$- 1 1 k 112 1 - k 1 = 0$

Expanding the Determinant

Expanding along the first row ( $R_{1}$ ):
$- 1 \cdot 12 - k 1 - 1 \cdot 1 k - k 1 + 1 \cdot 1 k 12 = 0$
Let's evaluate the first minor:
$- 1 \cdot (1 \cdot 1 - 2 \cdot (- k))$
$- 1 \cdot (1 + 2 k)$

Expanding the Second Term

Now for the second element of $R_{1}$ :
$- 1 \cdot 1 k - k 1$
$- 1 \cdot (1 \cdot 1 - k \cdot (- k))$
$- 1 \cdot (1 + k^{2})$

Expanding the Third Term

Finally, the third element of $R_{1}$ :
$+ 1 \cdot 1 k 12$
$+ 1 \cdot (1 \cdot 2 - k \cdot 1)$
$+ 1 \cdot (2 - k)$
Putting it all together:
$- (1 + 2 k) - (1 + k^{2}) + (2 - k) = 0$

Simplifying the Equation

Let's open the brackets and group like terms:
$- 1 - 2 k - 1 - k^{2} + 2 - k = 0$
Grouping $k^{2}$ terms: $- k^{2}$
Grouping $k$ terms: $- 2 k - k = - 3 k$
Grouping constants: $- 1 - 1 + 2 = 0$
The equation reduces to: $- k^{2} - 3 k = 0$

Solving the Quadratic Equation

Multiply by $- 1$ : $k^{2} + 3 k = 0$
Factor out $k$ :
$k (k + 3) = 0$
This gives two possible solutions:
$k = 0$ or $k + 3 = 0 ⟹ k = - 3$

Final Conclusion

The lines are coplanar for two specific values of $k$ .
$k = 0$ or $k = - 3$
This is a classic JEE problem testing the scalar triple product condition for 3D lines.
Correct Option: $k = 0$ or $- 3$

00:00 / 00:00

The Sigma Insight: Shortest Distance Between Two Skew Lines

Solution Diagram

Analyzing the Setup

Imagine you are standing in a vast, empty 3D room. Two straight lines are floating in front of you, suspended in mid-air. We want to determine if these lines are coplanar—that is, if they can be contained within a single, flat sheet of paper.

If the lines are skew, they exist in different "layers" of space and can never share a plane. To force these lines onto the same plane, we utilize the geometric property that the volume of the parallelepiped formed by their connecting vector and their direction vectors must be zero.

The Mathematical Bridge

We identify a point on the first line,

P_{1} (2, 3, 4)

, and a point on the second line,

P_{2} (1, 4, 5)

. We define the bridge vector connecting these points as

P_{1} P_{2} = (1 - 2, 4 - 3, 5 - 4) = (- 1, 1, 1)

The direction vectors of the two lines are given as

d_{1} = (1, 1, - k)

and

d_{2} = (k, 2, 1)

. For the lines to be coplanar, the scalar triple product of these three vectors must vanish:

[P_{1} P_{2} d_{1} d_{2}] = 0

The Determinant Dance

We assemble the components into a determinant and set it to zero:

- 1 1 k 112 1 - k 1 = 0

Expanding this determinant along the first row, we obtain:

- 1 \cdot 12 - k 1 - 1 \cdot 1 k - k 1 + 1 \cdot 1 k 12 = 0

Calculating each minor individually:

1. The first minor is

(1 - (- 2 k)) = (1 + 2 k)

. 2. The second minor is

(1 - (- k^{2})) = (1 + k^{2})

. 3. The third minor is

(2 - k)

The Collapse of Complexity

Substituting these values back into our equation, we get:

- 1 (1 + 2 k) - 1 (1 + k^{2}) + 1 (2 - k) = 0

Expanding the terms yields:

- 1 - 2 k - 1 - k^{2} + 2 - k = 0

The constant terms

- 1 - 1 + 2

cancel out perfectly. We are left with the simplified quadratic equation:

- k^{2} - 3 k = 0

Multiplying by

- 1

and factoring, we find:

k (k + 3) = 0

This yields the final solutions: $k = 0$ or $k = - 3$ . These values represent the specific conditions under which the two lines are coplanar.

Common Traps & FAQs

Wait, why do we use the scalar triple product? Can't we just find the intersection point?

That is a great question! Finding the intersection point is a valid strategy if you *know* the lines intersect. However, the problem asks for the condition of coplanarity, which includes both intersecting lines AND parallel lines. If the lines are parallel, they don't have a single intersection point, but they are still coplanar. The scalar triple product is a powerful tool because it covers both cases simultaneously—it equals zero if the lines are parallel (because the direction vectors are proportional) or if they intersect.

How do I know which point is

(x_{1}, y_{1}, z_{1})

and which is

(x_{2}, y_{2}, z_{2})

? Does the order matter?

It does not matter! If you swap the points, the vector

P_{1} P_{2}

becomes

P_{2} P_{1}

, which is just the negative of the original vector. Since the determinant is linear with respect to its rows, swapping the points only multiplies the entire determinant by

- 1

. Since we are setting the determinant to zero, multiplying by

- 1

doesn't change the final values of

k

at all. You are safe either way!

What if the lines are parallel? Does the formula still work?

Absolutely! If the lines are parallel, their direction vectors

d_{1}

and

d_{2}

are proportional. In a determinant, if one row is a scalar multiple of another, the determinant is automatically zero. So, the scalar triple product condition naturally accounts for the parallel case without you needing to check it separately. It is a very robust condition.

Animated Solution for Mathematics - Three Dimensional Geometry: 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

Select Answer:

Visualized Solution

Visualizing Coplanar Lines

The Condition for Coplanarity

Parameters of Line 1

Parameters of Line 2

Vector $P_{1} P_{2}$

Setting up the Determinant

Expanding the Determinant

Expanding the Second Term

Expanding the Third Term

Simplifying the Equation

Solving the Quadratic Equation

Final Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Mathematical Bridge

The Determinant Dance

The Collapse of Complexity

Similar Questions

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 ___

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

If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then $k$ is equal to:

If the straight lines $\frac{x - 1}{k} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 2}{3} = \frac{y - 3}{k} = \frac{z - 1}{2}$ intersect at a point, then the integer $k$ 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

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

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

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 :

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

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

Animated Solution for Mathematics - Three Dimensional Geometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The lines 1x−2​=1y−3​=−kz−4​ and kx−1​=2y−4​=1z−5​ are coplanar if

Select Answer:

Visualized Solution

Visualizing Coplanar Lines

The Condition for Coplanarity

Parameters of Line 1

Parameters of Line 2

Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​P2​​

Setting up the Determinant

Expanding the Determinant

Expanding the Second Term

Expanding the Third Term

Simplifying the Equation

Solving the Quadratic Equation

Final Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Mathematical Bridge

The Determinant Dance

The Collapse of Complexity

Similar Questions

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Animated Solution for Mathematics - Three Dimensional Geometry: 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

Vector $P_{1} P_{2}$

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 ___

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

If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then $k$ is equal to:

If the straight lines $\frac{x - 1}{k} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 2}{3} = \frac{y - 3}{k} = \frac{z - 1}{2}$ intersect at a point, then the integer $k$ 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

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

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

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 :

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

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