Q.Show that the lines $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other.

The key idea is that two lines are perpendicular if the dot product of their direction vectors is zero.

Step 1: Identify direction vectors from the symmetric equations.

For the first line, $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$, the direction vector is

$\vec{d_1} = (7, -5, 1)$.

Step 2: For the second line, $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$, the direction vector is

$\vec{d_2} = (1, 2, 3)$.

Step 3: Compute the dot product: …

The key to checking perpendicularity between lines in 3D is to look at their direction vectors. A line given in symmetric form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ has direction vector $(a, b, c)$. The actual points on the lines don't matter for perpendicularity — only the directions do.

1. Extract the direction vectors.

   For the first line: $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$, the denominators give $\vec{d_1} = (7, -5, 1)$.

   For the second line: $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$, the denominators give $\vec{d_2} = (1, 2, 3)$.

2. Apply the perpendicularity condition.

   Two lines are perpendicular if and only if the dot product of their direction vectors is zero: $\vec{d_1} \cdot \vec{d_2} = 0$. …

Method: Perpendicularity of lines given in symmetric form

Two lines are perpendicular exactly when their direction vectors have zero dot product. For lines written in symmetric form, the direction is read straight off the denominators.

Steps

Step 1: Read each direction vector from the denominators: $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$ gives $\vec d=(a,b,c)$. A denominator of $1$ (as under $z$) is a genuine component, not something to ignore.

Step 2: Compute the dot product.

$\vec d_1\cdot\vec d_2=a_1a_2+b_1b_2+c_1c_2.$ …

Common Mistakes

Mistake 1: Ignoring a denominator of $1$.

Why it's wrong: in $\frac{z}{1}$ the $z$-direction component is $1$, a real part of $\vec d_1=(7,-5,1)$; treating it as absent drops a term. Correct approach: include every denominator to get the full direction vector.

Mistake 2: Trying to show the lines meet. …