For real numbers α and β ≠ 0, if the point of intersection of the straight lines (x-α)/1 = (y-1)/2 =

Q: Wait, why do we equate the $y$ and $z$ coordinates first instead of the $x$ coordinates?

That is a brilliant observation! If we started by equating the $x$-coordinates, we would have an equation like $\lambda + \alpha = \beta\mu + 4$, which contains two unknowns ($\alpha$ and $\beta$) plus the parameters. By choosing the $y$ and $z$ coordinates, we isolate the parameters $\lambda$ and $\mu$ first. This is a classic strategy in 3D geometry: always look for the path of least resistance to find your parameters before tackling the constants.

Q: Can we use the plane equation before finding the intersection point?

Not directly. The plane equation is a constraint that applies specifically to the point of intersection $P$. You must first define the coordinates of $P$ using the line equations. Think of the lines as the 'search' for the point, and the plane as the 'filter' that confirms where that point must reside.

Q: Is it possible for the lines to be skew instead of intersecting?

In a general 3D geometry problem, yes, lines can be skew. However, this problem explicitly states that the lines intersect at a point $P$. In JEE Advanced problems, when the question provides a premise like 'the point of intersection... lies on the plane', you can be confident that the lines are indeed coplanar and intersecting. Trust the problem statement!

Question

For real numbers $\alpha$ and $\beta 
eq 0$, if the point of intersection of the straight lines $\frac{x-\alpha}{1} = \frac{y-1}{2} = \frac{z-1}{3}$ and $\frac{x-4}{\beta} = \frac{y-6}{3} = \frac{z-7}{3}$ lies on the plane $x + 2y - z = 8$, then $\alpha - \beta$ is equal to :

Options:

(A) 5

(B) 9

(C) 3

(D) 7

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional void. Two straight lines, $L_1$ and $L_2$, are streaking through this space. They are not parallel; they are destined to meet at a single, precise location: the point $P$.

Our mission is to find the values of the constants $\alpha$ and $\beta$ that define these lines, and ultimately, to calculate the difference $\alpha - \beta$. This is a journey of uncovering hidden constraints.

### The Parametric GPS

To navigate these lines, we use the parametric form. For line $L_1$, given by:

$$\frac{x-\alpha}{1} = \frac{y-1}{2} = \frac{z-1}{3} = \lambda$$

This transforms the line into a set of coordinates: $P = (\lambda + \alpha, 2\lambda + 1, 3\lambda + 1)$.

Similarly, for line $L_2$, defined by:

$$\frac{x-4}{\beta} = \frac{y-6}{3} = \frac{z-7}{3} = \mu$$

This gives us the general point $P = (\beta\mu + 4, 3\mu + 6, 3\mu + 7)$. We now have two different ways to describe the same point $P$ in space.

### The Intersection Logic

Since the lines intersect at $P$, the coordinates must be identical at that specific point. We look at the $y$ and $z$ components first, as they are free from the unknown constants $\alpha$ and $\beta$.

Equating the $y$-coordinates gives:

$$2\lambda + 1 = 3\mu + 6 \Rightarrow 2\lambda - 3\mu = 5$$

Equating the $z$-coordinates gives:

$$3\lambda + 1 = 3\mu + 7 \Rightarrow 3\lambda - 3\mu = 6$$

We now have a system of two linear equations. Subtracting the first from the second, the $\mu$ terms vanish, leaving us with $\lambda = 1$. Substituting this back, we find $\mu = -1$.

### The Plane as a Gatekeeper

Now that we know $\lambda = 1$, we can pin down the coordinates of $P$ using the $L_1$ parameterization:

$$P = (1 + \alpha, 2(1) + 1, 3(1) + 1) = (1 + \alpha, 3, 4)$$

The problem states that $P$ lies on the plane $x + 2y - z = 8$. Substituting our coordinates into the plane equation:

$$(1 + \alpha) + 2(3) - 4 = 8$$

Simplifying this, we get $1 + \alpha + 6 - 4 = 8$, which leads to $\alpha + 3 = 8$, or $\alpha = 5$.

### The Final Reveal

We return to the $x$-coordinates to solve for $\beta$. We know the $x$-coordinate of $P$ is $1 + \alpha = 6$. From the $L_2$ parameterization, the $x$-coordinate is $\beta\mu + 4$.

Substituting $\mu = -1$:

$$-\beta + 4 = 6 \Rightarrow \beta = -2$$

The final step is the subtraction:

$$\alpha - \beta = 5 - (-2) = 7$$

Through logical deduction and systematic elimination, we have navigated the 3D space and arrived at the solution: **7**.

Select Answer:

Visualized Solution

Visualizing the Geometry

Parametric Form of $L_{1}$

Parametric Form of $L_{2}$

Equating $y$ and $z$ Coordinates

Solving for $λ$

Solving for $μ$

Coordinates of Point $P$

Point $P$ on the Plane

Solving for $α$

Equating $x$ Coordinates for $β$

Solving for $β$

Final Calculation $α - β$

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Parametric GPS

The Intersection Logic

The Plane as a Gatekeeper

The Final Reveal

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Select Answer:

Visualized Solution

Visualizing the Geometry

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

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

Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z Coordinates

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

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

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

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

Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x Coordinates for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

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

Final Calculation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α−β