If the line, (x-3)/2 = (y+2)/-1 = (z+4)/3 lies in the plane, lx + my - z = 9, then l² + m² is equal

Q: Wait, why do we need two equations? Can't we just use the point on the line?

Great question! Using the point $(3, -2, -4)$ only gives us one constraint on the variables $l$ and $m$. Since we have two unknowns, we need two independent pieces of information. The second piece comes from the fact that the line's direction is perpendicular to the plane's normal.

Q: Why is the dot product of the line's direction and the plane's normal zero?

Think of the plane's normal vector as a needle sticking straight out of the surface. If the line lies flat on that surface, it is impossible for the line to have any component along that needle. Thus, they must be at a $90^\circ$ angle, making their dot product zero.

Q: How do I know which vector is the normal vector of the plane?

For any plane given by the equation $Ax + By + Cz = D$, the coefficients of $x, y,$ and $z$ form the components of the normal vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$. In your case, $l, m,$ and $-1$ are those coefficients.

Q: Is it possible for the line to be parallel to the plane but not lie in it?

Yes, absolutely! If the line were parallel but not in the plane, the dot product would still be zero, but the point on the line would not satisfy the plane's equation. The problem specifically states the line 'lies in' the plane, which forces both conditions to be true.

Q: What if I made a sign error in the dot product calculation?

Sign errors are the most common 'JEE traps'! Always write out the components clearly: $\vec{b} = (2, -1, 3)$ and $\vec{n} = (l, m, -1)$. Then, multiply component-wise: $(2)(l) + (-1)(m) + (3)(-1) = 0$. Taking that extra second to write it down prevents the most frustrating mistakes.

Question

If the line, $\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+4}{3}$ lies in the plane, $lx + my - z = 9$, then $l^2 + m^2$ is equal to :

Options:

(A) $5$

(B) $2$

(C) $26$

(D) $18$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty room. You have a thin, perfectly straight laser beam (our line) and a massive, flat sheet of glass (our plane). The problem asks us to consider a scenario where this laser beam is not just passing through the glass, but is resting perfectly flush against its surface.

This is the essence of the condition: **the line lies in the plane**.

### The Anchor Point

Every line in 3D space is defined by a direction and a starting point. Looking at our line equation,

$$\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+4}{3}$$

we can immediately extract a point that the line passes through. By setting the ratios to zero, we find the anchor point $P(3, -2, -4)$.

If the entire line is contained within the plane $lx + my - z = 9$, then this point $P$ must also be a point on the plane. It must satisfy the plane's equation. Substituting our coordinates, we get:

$$l(3) + m(-2) - (-4) = 9$$

Simplifying this, we arrive at our first vital constraint: $3l - 2m = 5$. This is our first **anchor equation**. We have successfully locked down one relationship between the variables $l$ and $m$.

### The Orthogonality of Directions

Now, let's think about the orientation. A plane is defined by its 'normal'—a vector $\vec{n}$ that points straight out from its surface. For our plane $lx + my - z = 9$, the normal vector is $\vec{n} = l\hat{i} + m\hat{j} - 1\hat{k}$.

Meanwhile, our line has a direction vector $\vec{b} = 2\hat{i} - 1\hat{j} + 3\hat{k}$. If the line is lying flat on the plane, it must be perpendicular to the normal vector. In the language of vectors, this means their dot product must vanish:

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

Calculating this, we get $(2)(l) + (-1)(m) + (3)(-1) = 0$, which simplifies beautifully to $2l - m = 3$. This is our second constraint. We have now captured the 'soul' of the geometry in two simple algebraic equations.

### The Elegant Resolution

We are left with a system of two linear equations:

1) $3l - 2m = 5$

2) $2l - m = 3$

From the second equation, we can easily see that $m = 2l - 3$. Substituting this into the first equation, we get:

$$3l - 2(2l - 3) = 5$$

Expanding this, we find $3l - 4l + 6 = 5$, which leads us to $-l = -1$, or simply $l = 1$. With $l$ in hand, finding $m$ is a breeze: $m = 2(1) - 3 = -1$.

### Final Calculation

We have found our variables! The final step is to calculate $l^2 + m^2$. Plugging in our values, we get:

$$(1)^2 + (-1)^2 = 1 + 1 = 2$$

The final result is **2**.

Animated Solution for Mathematics - Three Dimensional Geometry: If the line, $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z + 4}{3}$ lies in the plane, $l x + m y - z = 9$ , then $l^{2} + m^{2}$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Setup

Point on the Line

Satisfying the Plane Equation

Forming Equation (i)

Vectors of Line and Plane

Orthogonality Condition

Calculating the Dot Product

Forming Equation (ii)

Solving for l and m

Calculating l

Calculating m

Finding $l^{2} + m^{2}$

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Anchor Point

The Orthogonality of Directions

The Elegant Resolution

Final Calculation

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