The equation of the line passing through (-4, 3, 1), parallel to the plane x + 2y - z - 5 = 0 and in

Q: Wait, why do we assume the intersection point Q lies on L1?

The problem explicitly states that the new line must intersect the given line L1. By defining a general point Q on L1 using a parameter \lambda, we are essentially saying, 'If there is an intersection, it must be at this point for some value of \lambda.' This is a standard and powerful technique in 3D geometry to handle intersection conditions.

Q: Why is the dot product zero? I thought parallel meant the vectors are the same?

That is a common trap! If a line is parallel to a plane, its direction vector is perpendicular to the plane's normal vector. Think of the normal vector as a stick poking straight out of the plane. If the line is parallel to the plane, it will never hit that stick, meaning the line and the stick (the normal) are at a 90-degree angle. Hence, the dot product is zero.

Q: Can I use the cross product instead of the dot product?

The cross product is used to find a vector perpendicular to two other vectors. While you could use it to find the normal of a plane if you had two vectors on it, here we are using the dot product because it is the most direct way to enforce the perpendicularity condition between the line's direction vector and the plane's normal vector. Stick to the dot product for this; it is cleaner and faster!

Question

The equation of the line passing through $(-4, 3, 1)$, parallel to the plane $x + 2y - z - 5 = 0$ and intersecting the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z - 2}{-1}$ is:

Options:

(A) $\frac{x + 4}{-1} = \frac{y - 3}{1} = \frac{z - 1}{1}$

(B) $\frac{x + 4}{3} = \frac{y - 3}{-1} = \frac{z - 1}{1}$

(C) $\frac{x + 4}{1} = \frac{y - 3}{1} = \frac{z - 1}{3}$

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. You have a fixed point $A(-4, 3, 1)$ and a line $L_1$ defined by the symmetric equations:

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

You are tasked with finding a new line that passes through $A$, runs parallel to the plane $P: x + 2y - z - 5 = 0$, and intersects $L_1$. This might seem like a daunting puzzle, but let us break it down into a beautiful, logical sequence.

### The Power of Parametric Coordinates

To find our line, we need two things: a point on the line and its direction ratios. We already have the point $A$. The intersection point $Q$ is the key.

Since $Q$ lies on $L_1$, we can represent its coordinates using a single parameter $\lambda$. By setting:

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

We express the coordinates of $Q$ as $x = -3\lambda - 1$, $y = 2\lambda + 3$, and $z = -\lambda + 2$. Thus, $Q = (-3\lambda - 1, 2\lambda + 3, -\lambda + 2)$. This parameter $\lambda$ is our bridge to the solution.

### Defining the Direction Vector

Now that we have two points on our required line, $A$ and $Q$, we can easily determine the direction vector $\vec{d} = \vec{AQ}$. Subtracting the coordinates of $A$ from $Q$, we get:

$$
\vec{AQ} = (-3\lambda - 1 - (-4), 2\lambda + 3 - 3, -\lambda + 2 - 1)
$$

This simplifies to $\vec{AQ} = (3 - 3\lambda, 2\lambda, 1 - \lambda)$. This vector represents the orientation of our line in space, and it is currently waiting for us to find the correct value of $\lambda$.

### The Geometric Constraint

The problem states that our line is parallel to the plane $P$. The normal vector of the plane, $\vec{n}$, is derived from the coefficients of $x, y,$ and $z$ in the plane equation $x + 2y - z - 5 = 0$, giving us $\vec{n} = (1, 2, -1)$.

Because the line is parallel to the plane, its direction vector $\vec{AQ}$ must be perpendicular to the normal vector $\vec{n}$. This means their dot product must be zero: $\vec{AQ} \cdot \vec{n} = 0$.

Substituting our values, we get:

$$
(1)(3 - 3\lambda) + (2)(2\lambda) + (-1)(1 - \lambda) = 0
$$

Expanding this, we have $3 - 3\lambda + 4\lambda - 1 + \lambda = 0$, which simplifies to $2\lambda + 2 = 0$. Solving for $\lambda$, we find $\lambda = -1$.

### Final Calculation

With $\lambda = -1$, we can find the specific direction ratios of our line by substituting $\lambda$ back into $\vec{AQ}$. We get:

$$
\vec{AQ} = (3 - 3(-1), 2(-1), 1 - (-1)) = (6, -2, 2)
$$

To simplify, we can divide by $2$ to get the direction ratios $(3, -1, 1)$. Now, we have the point $A(-4, 3, 1)$ and the direction ratios $(3, -1, 1)$.

The symmetric equation of the line is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$. Substituting our values, we obtain the final result:

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

We have successfully navigated the 3D space to find the line that satisfies all conditions. Remember, geometry is not just about formulas; it is about visualizing the relationships between points, lines, and planes.

Animated Solution for Mathematics - Three Dimensional Geometry: The equation of the line passing through $(- 4, 3, 1)$ , parallel to the plane $x + 2 y - z - 5 = 0$ and intersecting the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ is:

Select Answer:

Visualized Solution

Visualizing the 3D Setup

General Point $Q$ on $L_{1}$

Direction Vector $A Q$

Parallelism Condition

Applying the Dot Product

Solving for $λ$

Finding Direction Ratios

Final Equation of the Line

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Power of Parametric Coordinates

Defining the Direction Vector

The Geometric Constraint

Final Calculation

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