The two lines x = ay + b, z = cy + d; and x = a'y + b', z = c'y + d' are perpendicular to each other

Q: Wait, why is the denominator for the $y$-term equal to 1? I don't see a denominator in the original equation.

That is a great observation! In the symmetric form, we want to express the line as $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$. Since our equations are $x = ay+b$ and $z = cy+d$, we can rearrange them to get $y = \frac{x-b}{a}$ and $y = \frac{z-d}{c}$. Since $y$ is just $y$, we can write it as $\frac{y-0}{1}$. Thus, the direction ratio corresponding to $y$ is indeed 1.

Q: Can I just use the coefficients $a$ and $c$ as the direction ratios directly without rearranging?

No, that is a common trap! The coefficients $a$ and $c$ are attached to $y$, but the standard symmetric form requires the variables $x, y, z$ to have coefficients of 1 in the numerator. By rearranging to isolate $y$, we ensure the denominators correctly represent the direction ratios. Always convert to symmetric form first!

Q: What if the lines were parallel instead of perpendicular?

If the lines were parallel, their direction vectors would be proportional, meaning $\frac{a}{a'} = \frac{1}{1} = \frac{c}{c'}$. This would imply $a = a'$ and $c = c'$. Perpendicularity, however, requires the dot product to be zero, which leads to the condition $aa' + 1 + cc' = 0$.

Question

The two lines $x = ay + b, z = cy + d$; and $x = a'y + b', z = c'y + d'$ are perpendicular to each other if

Options:

(A) $aa' + cc' = -1$

(B) $aa' + cc' = 1$

(C) $\frac{a}{a'} + \frac{c}{c'} = -1$

(D) $\frac{a}{a'} + \frac{c}{c'} = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional coordinate system. You see two lines stretching out into the void, defined by the parametric-like structure: $x = ay + b$ and $z = cy + d$.

At first glance, this might look intimidating, but it is actually a beautiful invitation to understand the true nature of direction in 3D space.

### The Illusion of Complexity

The equations $x = ay + b$ and $z = cy + d$ are not in the standard symmetric form we are accustomed to. In the standard symmetric form, a line is represented as:

$$\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$$

Here, $(l, m, n)$ represent the direction ratios. Our given equations imply that $y$ is the independent parameter that dictates the position of any point on the line.

To unlock the secrets of these lines, we must perform a simple algebraic transformation to reach the symmetric form.

### Extracting the DNA of the Line

Let us focus on the first line. From $x = ay + b$, we isolate $y$ to get $y = \frac{x-b}{a}$. From $z = cy + d$, we get $y = \frac{z-d}{c}$.

We can chain these together:

$$\frac{x-b}{a} = y = \frac{z-d}{c}$$

To align this with the standard symmetric form, we write $y$ as $\frac{y-0}{1}$. The equation becomes:

$$\frac{x-b}{a} = \frac{y-0}{1} = \frac{z-d}{c}$$

The denominators $(a, 1, c)$ are the direction ratios of the first line. This vector, $\vec{d}_1 = (a, 1, c)$, is the **DNA of the line**; it defines its orientation in space.

Repeating this process for the second line, $x = a'y + b'$ and $z = c'y + d'$, we find its direction vector: $\vec{d}_2 = (a', 1, c')$.

### The Beauty of Perpendicularity

We arrive at the heart of the problem: when are two lines perpendicular? Geometrically, this means the angle between them is $90^\circ$.

Algebraically, this requires the dot product of their direction vectors to be zero:

$$\vec{d}_1 \cdot \vec{d}_2 = 0$$

Substituting our vectors, we obtain:

$$(a)(a') + (1)(1) + (c)(c') = 0$$

Simplifying this, we find $aa' + 1 + cc' = 0$. Moving the constant to the other side, we reach the final condition.

### Final Result

The condition for the two lines to be perpendicular is:

**$aa' + cc' = -1$**

This is a simple, elegant result that emerges from the chaos of non-standard equations. The lesson here is clear: never be intimidated by the form of an equation, as algebraic manipulation will always reveal the underlying geometric truth.

Animated Solution for Mathematics - Three Dimensional Geometry: The two lines $x = a y + b, z = cy + d$ ; and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ are perpendicular to each other if

Select Answer:

Visualized Solution

Given 3D Lines

The Symmetric Form

Converting Line 1

Direction Ratios of Line 1

Converting Line 2

Direction Ratios of Line 2

Condition for Perpendicularity

Applying the Condition

Simplifying the Equation

Final Conclusion

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Illusion of Complexity

Extracting the DNA of the Line

The Beauty of Perpendicularity

Final Result

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Animated Solution for Mathematics - Three Dimensional Geometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The two lines x=ay+b,z=cy+d; and x=a′y+b′,z=c′y+d′ are perpendicular to each other if