Let A be a point on the line r = (1 - 3μ)i + (μ - 1)j + (2 + 5μ)k and B(3, 2, 6) be a point in the s

Q: Wait, why do we use the dot product? Shouldn't we use the cross product if they are parallel?

That is a classic confusion! If a *vector* is parallel to a *plane*, it means the vector lies within the plane or is parallel to it. Since the plane's normal vector is perpendicular to every line in the plane, our vector $\vec{AB}$ must be perpendicular to that normal vector. Perpendicularity is checked using the dot product, which must equal zero.

Q: Can I just substitute the coordinates of $A$ into the plane equation to find $\mu$?

No, that would only tell you if point $A$ lies *on* the plane. The problem asks for the condition where the *vector* $\vec{AB}$ is parallel to the plane, which is a different geometric constraint entirely. Always distinguish between a point being on a plane and a vector being parallel to a plane.

Q: How do I extract the normal vector from $x - 4y + 3z = 1$?

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, the coefficients are $1, -4,$ and $3$, so $\vec{n} = 1\hat{i} - 4\hat{j} + 3\hat{k}$.

Question

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

Options:

(A) $\frac{1}{2}$

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

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

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast 3D coordinate system. You have a fixed point $B$ at $(3, 2, 6)$ and a line defined by the parameter $\mu$, where a point $A$ is sliding along.

We want to find the exact moment when the vector $\vec{AB}$ is perfectly parallel to the plane $x - 4y + 3z = 1$. This is not just algebra; it is a dance of vectors in space.

### Pinning Down Point $A$

First, we must define our moving target. The line equation is given as $\vec{r} = (1 - 3\mu)\hat{i} + (\mu - 1)\hat{j} + (2 + 5\mu)\hat{k}$.

This means any point $A$ on this line has coordinates $A \equiv (1 - 3\mu, \mu - 1, 2 + 5\mu)$. Think of $\mu$ as time; as $\mu$ changes, $A$ traces out the line.

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

Now, we define the vector $\vec{AB}$ by finding the difference between the position vectors of $B$ and $A$. This is $\vec{AB} = \vec{B} - \vec{A}$.

Substituting our coordinates, we get:
$$\vec{AB} = (3 - (1 - 3\mu))\hat{i} + (2 - (\mu - 1))\hat{j} + (6 - (2 + 5\mu))\hat{k}$$

Simplifying this, we find:
$$\vec{AB} = (2 + 3\mu)\hat{i} + (3 - \mu)\hat{j} + (4 - 5\mu)\hat{k}$$

This vector represents the displacement from our moving point $A$ to our fixed destination $B$.

### The Geometric Insight

Here is the core of the problem. A vector is parallel to a plane if and only if it is perpendicular to the plane's normal vector.

The plane $x - 4y + 3z = 1$ has a normal vector $\vec{n} = 1\hat{i} - 4\hat{j} + 3\hat{k}$. If $\vec{AB}$ is parallel to the plane, then $\vec{AB} \perp \vec{n}$.

In the language of vectors, this means their dot product must be zero:
$$\vec{AB} \cdot \vec{n} = 0$$

### The Final Calculation

Let us perform the dot product:
$$(2 + 3\mu)(1) + (3 - \mu)(-4) + (4 - 5\mu)(3) = 0$$

Expanding this, we get:
$$(2 + 3\mu) - 12 + 4\mu + 12 - 15\mu = 0$$

Combining the terms, we have $(3\mu + 4\mu - 15\mu) + (2 - 12 + 12) = 0$, which simplifies to:
$$-8\mu + 2 = 0$$

Solving for $\mu$, we find $8\mu = 2$, or:
$$\mu = \frac{1}{4}$$

This is the magic value! At $\mu = \frac{1}{4}$, the vector $\vec{AB}$ aligns perfectly to be parallel to the plane. You have successfully navigated the 3D space and solved the puzzle.

Animated Solution for Mathematics - Three Dimensional Geometry: Let A be a point on the line $r = (1 - 3 μ) \hat{i} + (μ - 1) \hat{j} + (2 + 5 μ) \hat{k}$ and B(3, 2, 6) be a point in the space. Then the value of $μ$ for which the vector $A B$ is parallel to the plane $x - 4 y + 3 z = 1$ is :

Select Answer:

Visualized Solution

Visualizing the Geometry

Coordinates of Point $A$

Constructing Vector $A B$

Simplifying Vector $A B$

The Normal Vector

Condition for Parallelism

Setting up the Dot Product

Expanding the Equation

Solving for $μ$

Final Conclusion

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Pinning Down Point $A$

Constructing the Vector $A B$

The Geometric Insight

The Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Geometry

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

Constructing Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Simplifying Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB