Two lines $L_1 : x = 5, \frac{y}{3-\alpha} = \frac{z}{-2}$ and $L_2 : x = \alpha, \frac{y}{-1} = \frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value(s) Options: (A) 1 (B) 2 (C) 3 (D) 4

### Analyzing the Setup The essence of coplanarity is determining if two lines can lie on the same flat plane. For two lines $L_1$ and $L_2$ to be coplanar, they must either intersect or be parallel. We begin by translating the lines into the standard symmetric form: $$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$$ For $L_1$, the constraint $x=5$ implies the line is locked at $x=5$. We rewrite it as: $$\frac{x-5}{0} = \frac{y-0}{3-\alpha} = \frac{z-0}{-2}$$ From this, we identify the point $P_1(5, 0, 0)$ and the direction vector $\vec{d_1} = \langle 0, 3-\alpha, -2 \rangle$. Similarly, for $L_2$, we identify the point $P_2(\alpha, 0, 0)$ and the direction vector $\vec{d_2} = \langle 0, -1, 2-\alpha \rangle$. ### The Scalar Triple Product If these lines are coplanar, the vector connecting the two points, $\vec{P_1P_2}$, must lie in the same plane as the direction vectors $\vec{d_1}$ and $\vec{d_2}$. Geometrically, this means the volume of the parallelepiped formed by these three vectors must be zero. This condition is expressed by the scalar triple product: $$\vec{P_1P_2} \cdot (\vec{d_1} \times \vec{d_2}) = 0$$ We represent this as a determinant: $$\begin{vmatrix} \alpha-5 & 0 & 0 \\ 0 & 3-\alpha & -2 \\ 0 & -1 & 2-\alpha \end{vmatrix} = 0$$ ### The Algebraic Victory Expanding the determinant along the first row simplifies the calculation significantly: $$(\alpha-5) [ (3-\alpha)(2-\alpha) - (-1)(-2) ] = 0$$ Next, we simplify the expression inside the brackets: $$(\alpha^2 - 5\alpha + 6) - 2 = \alpha^2 - 5\alpha + 4$$ This leads us to the master equation: $$(\alpha-5)(\alpha^2 - 5\alpha + 4) = 0$$ Factoring the quadratic term, we obtain: $$(\alpha-5)(\alpha-4)(\alpha-1) = 0$$ The possible values for $\alpha$ are **$1, 4,$ and $5$**. Selecting the required values from the set, we conclude the solution is **$1$ and $4$**.

JEE Advanced 2013

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: Two lines $L_{1} : x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$ and $L_{2} : x = α, \frac{y}{- 1} = \frac{z}{2 - α}$ are coplanar. Then $α$ can take value(s)

Q: Wait, why is the x-component of the direction vector zero?

Great observation! When an equation is given as $x = 5$, it means $x$ is constant regardless of $y$ or $z$. In vector terms, the line has no 'velocity' or 'direction' along the x-axis. Thus, the x-component of the direction vector must be $0$.

Q: Do we need to check if the lines are parallel separately?

Not necessarily! The scalar triple product condition $\vec{P_1P_2} \cdot (\vec{d_1} \times \vec{d_2}) = 0$ is a universal condition. If the lines are parallel, $\vec{d_1} \times \vec{d_2}$ becomes the zero vector, making the product zero automatically. It covers all cases of coplanarity.

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Visualize the two lines as rigid rods floating in space. For them to be coplanar, they must either intersect or be parallel. The most powerful tool to check this is the scalar triple product of the vector connecting a point on each line and the two direction vectors.

Hint 2

Before jumping into the determinant, ensure you have extracted the direction vectors correctly. Remember, if a line is given as

x = 5

, it means the x-coordinate is constant, implying the x-component of the direction vector is

0

. Write both lines in the form

\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}

Hint 3

Construct the determinant using the coordinates of points

P_{1} (5, 0, 0)

and

P_{2} (α, 0, 0)

and the direction vectors

d_{1} = ⟨ 0, 3 - α, - 2 ⟩

and

d_{2} = ⟨ 0, - 1, 2 - α ⟩

. Set the determinant

α - 5 00 0 3 - α - 1 0 - 2 2 - α = 0

and solve for

α

Visualized Solution

Visualizing Coplanar Lines

Two lines are coplanar if they lie in the exact same plane.
Given lines:
$L_{1} : x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$
$L_{2} : x = α, \frac{y}{- 1} = \frac{z}{2 - α}$

Standard Symmetric Form

The standard symmetric form of a line is:
$\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}$
$(x_{1}, y_{1}, z_{1})$ is a point on the line.
$⟨ a, b, c ⟩$ is the direction vector.

Analyzing Line $L_{1}$

Rewrite $L_{1}$ : $\frac{x - 5}{0} = \frac{y - 0}{3 - α} = \frac{z - 0}{- 2}$
Point $P_{1} = (5, 0, 0)$
Direction vector $d_{1} = ⟨ 0, 3 - α, - 2 ⟩$

Analyzing Line $L_{2}$

Rewrite $L_{2}$ : $\frac{x - α}{0} = \frac{y - 0}{- 1} = \frac{z - 0}{2 - α}$
Point $P_{2} = (α, 0, 0)$
Direction vector $d_{2} = ⟨ 0, - 1, 2 - α ⟩$

The Coplanarity Condition

For two lines to be coplanar, the vector connecting their points ( $P_{1} P_{2}$ ) must lie in the same plane as $d_{1}$ and $d_{2}$ .
Mathematically, their scalar triple product is zero:
$P_{1} P_{2} \cdot (d_{1} \times d_{2}) = 0$

Setting up the Determinant

The scalar triple product 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 values:
$α - 5 00 0 3 - α - 1 0 - 2 2 - α = 0$

Expanding the Determinant

Expand along the first row ( $R_{1}$ ):
$(α - 5) [(3 - α) (2 - α) - (- 1) (- 2)] = 0$

Simplifying the Expression

Simplify inside the brackets:
$(α - 5) [(α^{2} - 5 α + 6) - 2] = 0$
Combine constant terms:
$(α - 5) (α^{2} - 5 α + 4) = 0$

Factoring the Quadratic

Factorize the quadratic term $α^{2} - 5 α + 4$ :
We need two numbers that multiply to $4$ and add to $- 5$ .
These numbers are $- 4$ and $- 1$ .
So, $α^{2} - 5 α + 4 = (α - 4) (α - 1)$

Solving for $α$

The complete factored equation is:
$(α - 5) (α - 4) (α - 1) = 0$
Setting each factor to zero gives the possible values:
$α = 1, 4, 5$
Comparing with the given options (1, 2, 3, 4), the valid answers are 1 and 4.

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The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

The essence of coplanarity is determining if two lines can lie on the same flat plane. For two lines

L_{1}

and

L_{2}

to be coplanar, they must either intersect or be parallel.

We begin by translating the lines into the standard symmetric form:

\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}

For

L_{1}

, the constraint

x = 5

implies the line is locked at

x = 5

. We rewrite it as:

\frac{x - 5}{0} = \frac{y - 0}{3 - α} = \frac{z - 0}{- 2}

From this, we identify the point

P_{1} (5, 0, 0)

and the direction vector

d_{1} = ⟨ 0, 3 - α, - 2 ⟩

. Similarly, for

L_{2}

, we identify the point

P_{2} (α, 0, 0)

and the direction vector

d_{2} = ⟨ 0, - 1, 2 - α ⟩

The Scalar Triple Product

If these lines are coplanar, the vector connecting the two points,

P_{1} P_{2}

, must lie in the same plane as the direction vectors

d_{1}

and

d_{2}

. Geometrically, this means the volume of the parallelepiped formed by these three vectors must be zero.

This condition is expressed by the scalar triple product:

P_{1} P_{2} \cdot (d_{1} \times d_{2}) = 0

We represent this as a determinant:

α - 5 00 0 3 - α - 1 0 - 2 2 - α = 0

The Algebraic Victory

Expanding the determinant along the first row simplifies the calculation significantly:

(α - 5) [(3 - α) (2 - α) - (- 1) (- 2)] = 0

Next, we simplify the expression inside the brackets:

(α^{2} - 5 α + 6) - 2 = α^{2} - 5 α + 4

This leads us to the master equation:

(α - 5) (α^{2} - 5 α + 4) = 0

Factoring the quadratic term, we obtain:

(α - 5) (α - 4) (α - 1) = 0

The possible values for

α

are $1, 4,$ and $5$ . Selecting the required values from the set, we conclude the solution is $1$ and $4$ .

Common Traps & FAQs

Wait, why is the x-component of the direction vector zero?

Great observation! When an equation is given as

x = 5

, it means

x

is constant regardless of

y

z

. In vector terms, the line has no 'velocity' or 'direction' along the x-axis. Thus, the x-component of the direction vector must be

0

Do we need to check if the lines are parallel separately?

Not necessarily! The scalar triple product condition

P_{1} P_{2} \cdot (d_{1} \times d_{2}) = 0

is a universal condition. If the lines are parallel,

d_{1} \times d_{2}

becomes the zero vector, making the product zero automatically. It covers all cases of coplanarity.

Why did we expand along the first row?

Because it's the most efficient path! The first row contains two zeros, which simplifies the expansion significantly. Always look for rows or columns with zeros to save time and reduce the chance of calculation errors.

Animated Solution for Mathematics - Three Dimensional Geometry: Two lines $L_{1} : x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$ and $L_{2} : x = α, \frac{y}{- 1} = \frac{z}{2 - α}$ are coplanar. Then $α$ can take value(s)

Select Answer:

Visualized Solution

Visualizing Coplanar Lines

Standard Symmetric Form

Analyzing Line $L_{1}$

Analyzing Line $L_{2}$

The Coplanarity Condition

Setting up the Determinant

Expanding the Determinant

Simplifying the Expression

Factoring the Quadratic

Solving for $α$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Scalar Triple Product

The Algebraic Victory

Similar Questions

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Select Answer:

Visualized Solution

Visualizing Coplanar Lines

Standard Symmetric Form

Analyzing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Analyzing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

The Coplanarity Condition

Setting up the Determinant

Expanding the Determinant

Simplifying the Expression

Factoring the Quadratic

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Scalar Triple Product

The Algebraic Victory

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