Suppose the line (x-2)/α = (y-2)/-5 = (z+2)/2 lies on the plane x + 3y - 2z + β = 0. Then (α + β) is

Q: Wait, why do we reject the negative root here? Or do we?

In this specific problem, we are solving linear equations for $\alpha$ and $\beta$. There are no quadratic terms, so we don't have to worry about rejecting roots. We simply solve the linear equations derived from the geometric conditions.

Q: Why can't we just use the distance formula to show the distance is zero?

You absolutely could! If the distance from any point on the line to the plane is zero, the line must lie on the plane. However, that approach is algebraically much more tedious. Using the point-satisfaction and perpendicularity conditions is the elegant, standard way to solve this in JEE Advanced.

Q: How do I know which point to pick from the line equation?

The symmetric form of a line is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. The point $(x_1, y_1, z_1)$ is explicitly visible in the numerators. Just be careful with the signs: if you see $z+2$, that is $z - (-2)$, so the z-coordinate is $-2$.

Question

Suppose the line $\frac{x-2}{\alpha} = \frac{y-2}{-5} = \frac{z+2}{2}$ lies on the plane $x + 3y - 2z + \beta = 0$. Then $(\alpha + \beta)$ is equal to .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You have a perfectly flat sheet of paper, representing a plane, and you place a straight pen, representing a line, so that it rests completely flat on that paper.

We are given a line $L$ defined by:

$$ \frac{x-2}{\alpha} = \frac{y-2}{-5} = \frac{z+2}{2} $$

And a plane $P$ defined by:

$$ x + 3y - 2z + \beta = 0 $$

The problem states that the line lies entirely on the plane. This is a geometric constraint that unlocks two powerful mathematical keys.

### Phase 1: The Pothole Analogy

Think of the line as a road and the plane as a city. If the entire road is in the city, then every single point on that road must also be in the city.

If the line lies on the plane, every point on that line must satisfy the equation of the plane. By looking at the symmetric form of the line, we can identify a point $P$ that the line passes through.

By setting the ratios to zero, we find the point $P = (2, 2, -2)$. Since this point $P$ is on the line, and the line is on the plane, $P$ must satisfy the plane's equation:

$$ x + 3y - 2z + \beta = 0 $$

Substituting $x=2$, $y=2$, and $z=-2$, we get:

$$ (2) + 3(2) - 2(-2) + \beta = 0 $$

Expanding this, we have $2 + 6 + 4 + \beta = 0$, which simplifies to $12 + \beta = 0$. Thus, we find our first treasure: **$\beta = -12$**.

### Phase 2: The Vector Dance

Now, we need to find $\alpha$. We have the point, but we must consider the orientation.

The line has a direction vector $\vec{d} = (\alpha, -5, 2)$, which we extract from the denominators. The plane has a normal vector $\vec{n} = (1, 3, -2)$, which we extract from the coefficients of $x, y,$ and $z$.

If the line is lying flat on the plane, its direction vector must be perpendicular to the plane's normal vector. Mathematically, this means their dot product must be zero:

$$ \vec{d} \cdot \vec{n} = 0 $$

Let's perform the calculation:

$$ (\alpha)(1) + (-5)(3) + (2)(-2) = 0 $$

This gives us $\alpha - 15 - 4 = 0$, or $\alpha - 19 = 0$. Solving this, we get **$\alpha = 19$**.

### Final Calculation

We have navigated the geometry and unlocked the algebra. We found $\beta = -12$ and $\alpha = 19$.

The question asks for the sum $(\alpha + \beta)$. Substituting our values:

$$ 19 + (-12) = 7 $$

The final result is **$7$**.

Animated Solution for Mathematics - Three Dimensional Geometry: Suppose the line $\frac{x - 2}{α} = \frac{y - 2}{- 5} = \frac{z + 2}{2}$ lies on the plane $x + 3 y - 2 z + β = 0$ . Then $(α + β)$ is equal to .

Enter Numerical Value:

Visualized Solution

The Geometric Setup

Extracting a Point from the Line

The Point Satisfies the Plane

Substituting $P (2, 2, - 2)$ into Plane $P$

Calculating the Value of $β$

Direction and Normal Vectors

Applying the Perpendicularity Condition

Expanding the Dot Product

Calculating the Value of $α$

Finding $(α + β)$

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Phase 1

Phase 2

Final Calculation

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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; } Suppose the line αx−2​=−5y−2​=2z+2​ lies on the plane x+3y−2z+β=0. Then (α+β) is equal to .