Let P(3, 2, 6) be a point in space and Q be a point on the line r = (i - j + 2k) + μ(-3i + j + 5k).

Q: Wait, why do we use the dot product with the normal vector? Shouldn't we use the plane equation directly?

That is a fantastic question! The plane equation $x - 4y + 3z = 1$ defines the orientation of the plane in space. A vector is parallel to a plane if and only if it is perpendicular to the normal vector of that plane. By taking the dot product with the normal vector $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$, we are essentially checking if the vector $\vec{PQ}$ lies in a direction that never 'pierces' the plane, which is exactly what being parallel means.

Q: Why is the normal vector $\vec{n} = (1, -4, 3)$?

In the general equation of a plane $Ax + By + Cz = D$, the coefficients of $x, y,$ and $z$ represent the components of the normal vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$. This vector is always perpendicular to any line or vector lying within that plane.

Q: Can I just plug the coordinates of $Q$ into the plane equation?

No, that would only be correct if point $Q$ were required to lie *on* the plane. Here, we are looking for a specific position of $Q$ on the line such that the vector $\vec{PQ}$ is parallel to the plane. The distance from the plane doesn't have to be zero; the orientation of the vector is what matters.

Q: What if the dot product equals something other than zero?

If the dot product were non-zero, it would mean the vector $\vec{PQ}$ has a component along the normal vector, meaning it is 'tilted' relative to the plane and would eventually intersect it. Only when the dot product is zero is the vector perfectly 'flat' relative to the plane's surface.

Q: Is there any ambiguity about the direction of the normal vector?

Not at all! Whether you use $\vec{n}$ or $-\vec{n}$, the condition $\vec{PQ} \cdot \vec{n} = 0$ remains the same. The sign of the normal vector does not change the orthogonality condition.

Question

Let $P(3, 2, 6)$ be a point in space and $Q$ be a point on the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$. Then the value of $\mu$ for which the vector $\vec{PQ}$ is parallel to the plane $x - 4y + 3z = 1$ is

Options:

(A) $\frac{1}{4}$

(B) $-\frac{1}{4}$

(C) $\frac{1}{8}$

(D) $-\frac{1}{8}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. You have a fixed point $P(3, 2, 6)$ suspended in the air.

Nearby, there is a line defined by the equation $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$. This line acts as a trajectory, and point $Q$ is a traveler moving along this path, its position dictated by the parameter $\mu$.

Our goal is to find the exact value of $\mu$ where the vector connecting our fixed point $P$ to the traveler $Q$ becomes perfectly parallel to the plane $x - 4y + 3z = 1$.

### Defining the Traveler

Before we can solve the problem, we must define the position of $Q$. Since $Q$ lies on the line, its coordinates must satisfy the line equation.

By expanding the vector form, we can write the coordinates of $Q$ as a function of $\mu$:

$$Q = (1 - 3\mu, -1 + \mu, 2 + 5\mu)$$

This expression is the heartbeat of our problem. Every value of $\mu$ yields a unique point $Q$, and consequently, a unique vector $\vec{PQ}$.

### Constructing the Vector $\vec{PQ}$

Now, let us determine the vector $\vec{PQ}$. This vector represents the displacement from $P$ to $Q$. We subtract the coordinates of $P(3, 2, 6)$ from the coordinates of $Q$:

$$\vec{PQ} = (1 - 3\mu - 3)\hat{i} + (-1 + \mu - 2)\hat{j} + (2 + 5\mu - 6)\hat{k}$$

Simplifying this expression, we obtain:

$$\vec{PQ} = (-2 - 3\mu)\hat{i} + (\mu - 3)\hat{j} + (5\mu - 4)\hat{k}$$

This vector is our "arrow" in space. We require this arrow to be parallel to the plane $x - 4y + 3z = 1$.

### The Geometric Secret

A plane is defined by its normal vector—a vector that stands perfectly perpendicular to its surface. For our plane $x - 4y + 3z = 1$, the normal vector is $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$.

If our vector $\vec{PQ}$ is parallel to the plane, it must be perpendicular to the plane's normal vector. In the language of mathematics, this implies that their dot product must be zero:

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

### The Final Calculation

Now, we perform the dot product by multiplying the corresponding components of $\vec{PQ}$ and $\vec{n}$ and summing them:

$$1(-2 - 3\mu) - 4(\mu - 3) + 3(5\mu - 4) = 0$$

Expanding this carefully, we have:

$$-2 - 3\mu - 4\mu + 12 + 15\mu - 12 = 0$$

Grouping the terms involving $\mu$ and the constants, we find:

$$(-3\mu - 4\mu + 15\mu) + (-2 + 12 - 12) = 0$$

This simplifies to:

$$8\mu - 2 = 0$$

Solving for $\mu$, we arrive at the final result:

$$\mu = \frac{2}{8} = \frac{1}{4}$$

### Conclusion

At **$\mu = \frac{1}{4}$**, the vector $\vec{PQ}$ aligns perfectly parallel to the plane. It is a moment of mathematical harmony where the line, the point, and the plane all find their balance.

Remember, in JEE Advanced, success is rarely about brute force; it is about identifying the geometric condition—the "key"—that unlocks the equation. You have successfully applied the principles of spatial reasoning to solve this problem.

Select Answer:

Visualized Solution

Visualizing the Setup

Coordinates of Point $Q$

Constructing Vector $P Q$

Simplifying Vector $P Q$

Identifying the Plane's Normal

The Geometric Condition

Setting up the Dot Product

Expanding the Terms

Combining Like Terms

Final Conclusion

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Defining the Traveler

Constructing the Vector $P Q$

The Geometric Secret

The Final Calculation

Conclusion

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