A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x +

Q: Why do we use two different parameters, $\lambda$ and $\mu$, instead of just one?

Great question! The intersection points $P$ and $Q$ are independent of each other. If we used the same parameter, we would be forcing the points to be at the same 'relative' position on both lines, which is not necessarily true. By using $\lambda$ and $\mu$, we allow the math to find the specific, independent locations that satisfy the geometric constraint of the transversal line.

Q: How do I know if the lines are skew or intersecting?

In this problem, the existence of a transversal line with specific direction ratios implies that the lines are skew. If they were intersecting, the transversal would just be the line passing through their point of intersection, and the problem would be trivial. The beauty of this method is that it works regardless of whether the lines are skew or intersecting!

Q: Why can't we just use the vector equation of the lines?

You absolutely can! Using vector notation $\vec{r} = \vec{a} + t\vec{b}$ is often more elegant. However, for coordinate geometry problems like this, converting to Cartesian form and using parameters is often more direct for finding specific intersection points. Both methods lead to the same result; choose the one that feels more intuitive to you.

Q: What if I get a contradiction when solving for $\lambda$ and $\mu$?

If you encounter a contradiction (like $0 = 1$), it usually means the lines are parallel or the transversal doesn't exist as described. In JEE Advanced problems, if a transversal is guaranteed, the system will always be consistent. Double-check your subtraction of coordinates—that is where most sign errors hide!

Question

A line with direction cosines proportional to 2, 1, 2 meets each of the lines $x = y + a = z$ and $x + a = 2y = 2z$. The co-ordinates of each of the points of intersection are given by

Options:

(A) $(2a, 3a, 3a), (2a, a, a)$

(B) $(3a, 2a, 3a), (a, a, a)$

(C) $(3a, 2a, 3a), (a, a, 2a)$

(D) $(a, 3a, 3a), (a, a, a)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional void. You see two lines, $L_1$ and $L_2$, which are skew lines—they are neither parallel nor intersecting.

A third line, acting as a bridge, cuts through this space to touch both $L_1$ and $L_2$. We are given that the direction ratios of this bridge are $(2, 1, 2)$.

Our mission is to find the exact coordinates where this bridge meets the two lines.

### Phase 1: The Power of Parameterization

To find these points, we use the language of parameters. For line $L_1$, defined by $x = y + a = z$, we set the expression equal to a scalar $\lambda$.

This transforms the line into a set of coordinates: $x = \lambda$, $y = \lambda - a$, and $z = \lambda$. Thus, any point $P$ on $L_1$ is described as $(\lambda, \lambda - a, \lambda)$.

For line $L_2$, defined by $x + a = 2y = 2z$, we set the expression equal to a new parameter $\mu$. This yields:

$$x = \mu - a, \quad y = \frac{\mu}{2}, \quad z = \frac{\mu}{2}$$

Consequently, any point $Q$ on $L_2$ is $(\mu - a, \frac{\mu}{2}, \frac{\mu}{2})$.

### Phase 2: The Bridge of Direction Ratios

We now examine the line segment $PQ$. The direction ratios of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by the differences $(x_1 - x_2, y_1 - y_2, z_1 - z_2)$.

By subtracting the coordinates of $Q$ from $P$, we obtain the direction ratios of our transversal:

$$(\lambda - (\mu - a), (\lambda - a) - \frac{\mu}{2}, \lambda - \frac{\mu}{2})$$

This simplifies to the vector $(\lambda - \mu + a, \lambda - a - \frac{\mu}{2}, \lambda - \frac{\mu}{2})$. This vector represents the orientation of our bridge in space.

### Phase 3: The Proportionality Constraint

The problem states that the bridge has direction ratios proportional to $(2, 1, 2)$. Mathematically, this implies:

$$\frac{\lambda - \mu + a}{2} = \frac{\lambda - a - \frac{\mu}{2}}{1} = \frac{\lambda - \frac{\mu}{2}}{2}$$

We solve this by first equating the first and third ratios, which share the same denominator:

$$\lambda - \mu + a = \lambda - \frac{\mu}{2}$$

The $\lambda$ terms cancel out, leaving $a = \frac{\mu}{2}$, which simplifies to **$\mu = 2a$**.

Next, we equate the second and third ratios:

$$\frac{\lambda - a - \frac{\mu}{2}}{1} = \frac{\lambda - \frac{\mu}{2}}{2}$$

Substituting $\mu = 2a$ into this equation, we get:

$$\lambda - 2a = \frac{\lambda - a}{2}$$

Solving for $\lambda$, we find **$\lambda = 3a$**.

### The Final Reveal

We now substitute our parameters back into the coordinate expressions. For point $P$ on $L_1$:

$$P = (3a, 3a - a, 3a) = (3a, 2a, 3a)$$

For point $Q$ on $L_2$:

$$Q = (2a - a, \frac{2a}{2}, \frac{2a}{2}) = (a, a, a)$$

We have successfully bridged the gap. The transversal meets $L_1$ at **$(3a, 2a, 3a)$** and $L_2$ at **$(a, a, a)$**.

Animated Solution for Mathematics - Three Dimensional Geometry: A line with direction cosines proportional to 2, 1, 2 meets each of the lines $x = y + a = z$ and $x + a = 2 y = 2 z$ . The co-ordinates of each of the points of intersection are given by

Select Answer:

Visualized Solution

Visualizing the Setup

Parametrizing Point $P$ on $L_{1}$

Parametrizing Point $Q$ on $L_{2}$

Direction Ratios of $P Q$

Applying Proportionality Condition

Solving for $μ$

Finding the Value of $μ$

Solving for $λ$

Finding the Value of $λ$

Final Coordinates of Point $P$

Final Coordinates of Point $Q$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Final Reveal

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Select Answer:

Visualized Solution

Visualizing the Setup

Parametrizing Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Parametrizing Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Direction Ratios of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ