Let P be a plane containing the line (x-1)/3 = (y+6)/4 = (z+5)/2 and parallel to the line (x-3)/4 =

Q: Wait, why can't we just use the point $(3, 2, -5)$ from the second line as our point on the plane?

That is a classic trap! The problem states the plane is *parallel* to the second line, not that it *contains* it. A point on a parallel line does not necessarily lie on the plane itself. Always ensure the point you use is explicitly on the plane (like any point on the line $L_1$ which is contained in the plane).

Q: Does the order of the cross product matter? Like $\vec{d_1} \times \vec{d_2}$ vs $\vec{d_2} \times \vec{d_1}$?

Great question! Mathematically, $\vec{d_2} \times \vec{d_1} = -(\vec{d_1} \times \vec{d_2})$. This just flips the direction of your normal vector. Since the plane equation $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ is homogeneous, multiplying the entire equation by $-1$ gives you the exact same plane. So, the order doesn't change the final result!

Q: How do I know which point to pick from the first line?

Any point on the line $L_1$ will work! Since the line is contained in the plane, every single point on that line is also a point on the plane. The simplest way is to set the ratios equal to a parameter (like $k=0$) to find a convenient point, such as $(1, -6, -5)$.

Question

Let $P$ be a plane containing the line $\frac{x-1}{3} = \frac{y+6}{4} = \frac{z+5}{2}$ and parallel to the line $\frac{x-3}{4} = \frac{y-2}{-3} = \frac{z+5}{7}$. If the point $(1, -1, \alpha)$ lies on the plane $P$, then the value of $|5\alpha|$ is equal to ___

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler in the realm of 3D geometry. Today, we are not just solving an equation; we are constructing a plane in space.

Imagine you are standing in a vast, empty room. You have a straight, rigid rod (our line $L_1$) that is resting perfectly on a flat, infinite sheet of paper (our plane $P$).

You have another rod (our line $L_2$) floating in the air, perfectly parallel to that sheet of paper. Your mission is to define the orientation of that sheet of paper using nothing but the directions of these two rods.

### Extracting the DNA of the Lines

Every line in 3D space carries its identity in its direction vector. Look at the equation of our first line:

$$\frac{x-1}{3} = \frac{y+6}{4} = \frac{z+5}{2}$$

The denominators are the components of its direction vector, $\vec{d_1} = 3\hat{i} + 4\hat{j} + 2\hat{k}$. This vector tells us exactly how the line is tilted in space.

Similarly, for the second line, $\frac{x-3}{4} = \frac{y-2}{-3} = \frac{z+5}{7}$, we extract $\vec{d_2} = 4\hat{i} - 3\hat{j} + 7\hat{k}$. These two vectors are the keys to our kingdom.

### The Magic of the Cross Product

Now, we need the normal vector $\vec{n}$ to our plane. Think of the normal vector as the 'spine' of the plane—it is perpendicular to every single line lying on that surface.

Since our plane contains $L_1$, the vector $\vec{d_1}$ must be perpendicular to $\vec{n}$. Since the plane is parallel to $L_2$, the vector $\vec{d_2}$ must also be perpendicular to $\vec{n}$.

We need a vector that is simultaneously perpendicular to both $\vec{d_1}$ and $\vec{d_2}$. The cross product is our perfect tool for this: $\vec{n} = \vec{d_1} 	imes \vec{d_2}$.

We set up the determinant:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 2 \ 4 & -3 & 7 \end{vmatrix}$$

Expanding this, we get $\vec{n} = \hat{i}(28 - (-6)) - \hat{j}(21 - 8) + \hat{k}(-9 - 16)$, which simplifies to $\vec{n} = 34\hat{i} - 13\hat{j} - 25\hat{k}$. This vector $\vec{n} = (34, -13, -25)$ is the orientation of our plane.

### Building the Equation

We have the orientation, but we need a location. We need a point on the plane.

Since the plane contains $L_1$, any point on $L_1$ is a point on the plane. From the equation $\frac{x-1}{3} = \frac{y+6}{4} = \frac{z+5}{2}$, we can easily identify the point $A(1, -6, -5)$.

Now, we use the standard point-normal form: $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$. Substituting our normal vector $(34, -13, -25)$ and our point $(1, -6, -5)$, we get:

$$34(x-1) - 13(y+6) - 25(z+5) = 0$$

### The Final Reveal

We are told that a point $Q(1, -1, \alpha)$ lies on this plane. This is the final test.

If $Q$ is on the plane, it must satisfy the equation we just built. Let's substitute $x=1, y=-1, z=\alpha$:

$$34(1-1) - 13(-1+6) - 25(\alpha+5) = 0$$

This simplifies beautifully: $34(0) - 13(5) - 25(\alpha+5) = 0$, which becomes $-65 - 25\alpha - 125 = 0$.

Combining the constants, we get $-190 - 25\alpha = 0$, or $25\alpha = -190$. Dividing by $5$, we find $5\alpha = -38$.

The question asks for the absolute value $|5\alpha|$, which is $|-38| = \mathbf{38}$.

Enter Numerical Value:

Visualized Solution

Geometric Setup

Direction Vectors $d_{1}$ and $d_{2}$

Normal Vector $n$

Cross Product Setup

Evaluating $n$

Point on Plane $P$

Equation of Plane $P$

Substituting Point $Q$

Simplifying the Equation

Solving for $5 α$

Final Answer

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Extracting the DNA of the Lines

The Magic of the Cross Product

Building the Equation

The Final Reveal

Similar Questions

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If a plane passes through the points $(- 1, k, 0)$ , $(2, k, - 1)$ , $(1, 1, 2)$ and is parallel to the line $\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$ , then the value of $\frac{k ^{2} + 1}{( k - 1 ) ( k - 2 )}$ is

Enter Numerical Value:

Visualized Solution

Geometric Setup

Direction Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d1​​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2​​

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