The value of k such that (x-4)/1 = (y-2)/1 = (z-k)/2 lies in the plane 2x - 4y + z = 7, is
(a) 7
(b)

Q: Wait, why do we need to check the dot product? Can't we just plug in the point $(4, 2, k)$ and solve for $k$?

That is a common trap! If you only plug in the point, you are ensuring that the line *intersects* the plane at that point. However, the line could be piercing through the plane at an angle. To ensure the line *lies entirely within* the plane, it must be parallel to the plane first. The dot product condition $\vec{d} \cdot \vec{n} = 0$ is what guarantees that parallelism.

Q: How do I know which point to pick from the symmetric form of the line?

The symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ is designed to reveal a point on the line directly. By comparing your equation $\frac{x-4}{1} = \frac{y-2}{1} = \frac{z-k}{2}$ to the standard form, you can see that $x_1=4$, $y_1=2$, and $z_1=k$. Thus, the point is $(4, 2, k)$.

Q: What if the dot product $\vec{d} \cdot \vec{n}$ was not zero? Does that mean there is no solution?

Exactly! If the dot product is not zero, the line is not parallel to the plane. It would intersect the plane at exactly one point. If the line is not parallel, it can never lie entirely within the plane, no matter what value you choose for $k$. In such a case, the answer would indeed be 'no real value'.

Question

The value of $k$ such that $\frac{x-4}{1} = \frac{y-2}{1} = \frac{z-k}{2}$ lies in the plane $2x - 4y + z = 7$, is

Options:

(A) $7$

(B) $-7$

(C) no real value

(D) $4$

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, which we call a plane, defined by the equation $2x - 4y + z = 7$.

Now, imagine a thin, infinite wire, which we call a line, defined by the symmetric equations:

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

The problem asks us to find the value of $k$ that forces this wire to lie perfectly flat, embedded within that sheet of paper. This is not just an algebraic exercise; it is a beautiful dance of vectors and coordinates.

### The Two Pillars of Containment

To solve this, we need to satisfy two fundamental geometric conditions. First, the line must be parallel to the plane. If the line is tilted even slightly, it will eventually pierce through the paper, and it cannot lie entirely within it.

Second, at least one point on that line must be touching the plane. If the line is parallel but hovering above or below the plane, it is not 'in' the plane. Together, these two conditions lock the line into place.

### The Toolkit: Vectors

Let us extract the information we need. The line gives us a direction vector, $\vec{d}$, which tells us where it is heading. Looking at the denominators of our symmetric equation, we see $\vec{d} = 1\hat{i} + 1\hat{j} + 2\hat{k}$.

The plane gives us a normal vector, $\vec{n}$, which points directly away from its surface. From the coefficients of the plane equation $2x - 4y + z = 7$, we identify $\vec{n} = 2\hat{i} - 4\hat{j} + 1\hat{k}$.

Now, for the first condition: parallelism. If the line is parallel to the plane, its direction vector $\vec{d}$ must be perpendicular to the plane's normal vector $\vec{n}$.

We test this with the dot product:

$$\vec{d} \cdot \vec{n} = (1)(2) + (1)(-4) + (2)(1)$$

Calculating this, we get $2 - 4 + 2 = 0$. The dot product is zero! The vectors are orthogonal, and the line is indeed parallel to the plane. The geometry is sound.

### The Final Execution

With parallelism confirmed, we move to the second pillar: point containment. We know that the point $P(4, 2, k)$ lies on our line.

If the entire line is to lie within the plane, this specific point $P$ must also satisfy the plane's equation. We substitute the coordinates of $P$ into the plane equation $2x - 4y + z = 7$:

$$2(4) - 4(2) + (k) = 7$$

Let us perform the arithmetic with care. We have $8 - 8 + k = 7$.

The $8$ and $-8$ cancel out with elegant simplicity, leaving us with the final result: **$k = 7$**. This is the magic of the problem—the variables align, the constants vanish, and we are left with the precise value that anchors our line into the plane.

Animated Solution for Mathematics - Three Dimensional Geometry: The value of $k$ such that $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies in the plane $2 x - 4 y + z = 7$ , is

Select Answer:

Visualized Solution

Visualizing the Line and Plane

Condition 1: The Point Lies on the Plane

Condition 2: Orthogonality of Vectors

Extracting Point $P$ from the Line

Extracting Vectors $d$ and $n$

Checking $d \cdot n = 0$

Substituting $P$ into the Plane Equation

Solving the Linear Equation

Final Conclusion

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Two Pillars of Containment

The Toolkit

The Final Execution

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