Question

The equation of line which is parallel to $\frac{2-x}{-3}=\frac{y-2}{2}=\frac{z-4}{1}$ and passing through the point $(1,1,1)$, is

• A. $\frac{x-1}{3}=\frac{y-1}{2}=\frac{z+1}{1}$
• B. $\frac{x-1}{3}=\frac{y-1}{2}=\frac{z-1}{1}$
• C. $\frac{x-1}{3}=\frac{y-1}{-2}=\frac{z-1}{1}$
• D. $\frac{x-1}{-3}=\frac{y-1}{2}=\frac{z-1}{1}$
• E. $\frac{x-1}{3}=\frac{y-1}{2}=\frac{z+1}{-1}$

Hint:

Logic Tip: The most common trap in 3D geometry is failing to notice reversed signs in the numerator like $(2-x)$ or $(3-2y)$. Always force the variable to have a coefficient of $+1$ before reading the direction ratios.

Answer 1

The Correct Option is B

Concept:
The standard Cartesian form of a line is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, where $(x_1, y_1, z_1)$ is a point on the line and $\langle a, b, c \rangle$ are its direction ratios. Two lines are parallel if their direction ratios are proportional (or identical). It is crucial that the numerators are exactly in the form $x - x_1$, $y - y_1$, etc.
Step 1: Standardize the given line equation.
The given equation is $\frac{2-x}{-3}=\frac{y-2}{2}=\frac{z-4}{1}$. The first term is written as $2-x$. We must multiply both the numerator and denominator by $-1$ to put it into the standard $x - x_1$ form: $$\frac{-(2-x)}{-(-3)} = \frac{x-2}{3}$$ So the standardized line is: $$\frac{x-2}{3} = \frac{y-2}{2} = \frac{z-4}{1}$$
Step 2: Extract the direction ratios.
From the standardized equation, the direction ratios $\langle a, b, c \rangle$ are the denominators: $$\langle 3, 2, 1 \rangle$$
Step 3: Construct the equation of the new line.
The new line is parallel, so it uses the same direction ratios $\langle 3, 2, 1 \rangle$. It passes through the point $(x_1, y_1, z_1) = (1, 1, 1)$. Substitute these into the standard Cartesian formula: $$\frac{x - 1}{3} = \frac{y - 1}{2} = \frac{z - 1}{1}$$