Question:

The plane $ x - 2y + z = 0 $ is parallel to the line

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To identify if two lines are parallel, check if their direction ratios are proportional. The general form of the equation for a line is $ \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} $, where $ (a, b, c) $ is the direction ratio.

Updated On: May 2, 2025

$ \frac{x-3}{4} = \frac{y-4}{5} = \frac{z-3}{6} $
$ \frac{x-2}{4} = \frac{y-7}{5} = \frac{z-3}{7} $
$ \frac{x-2}{3} = \frac{y-3}{3} = \frac{z-4}{4} $
$ \frac{x-4}{3} = \frac{y-5}{4} = \frac{z-6}{3} $

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The Correct Option is A

Approach Solution - 1

To determine which line is parallel to the plane given by the equation $x - 2y + z = 0$, we must identify the direction vector normal to the plane and compare it with the direction vectors of the given lines.

The general form of a plane is $ax + by + cz = d$, where $(a, b, c)$ is the normal vector. Therefore, the normal vector to the plane $x - 2y + z = 0$ is $(1, -2, 1)$.

For a line given in the form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, the direction vector is $(a, b, c)$.

We need to find a line that has a direction vector orthogonal to the normal vector $(1, -2, 1)$. Two vectors $(a, b, c)$ and $(x, y, z)$ are orthogonal if their dot product is zero: $ax + by + cz = 0$.

Evaluating each option:

$\frac{x-3}{4} = \frac{y-4}{5} = \frac{z-3}{6}$ has direction vector $ (4, 5, 6) $. Calculate $1 \cdot 4 + (-2) \cdot 5 + 1 \cdot 6 = 4 - 10 + 6 = 0$. They are orthogonal, so the line is parallel.
$\frac{x-2}{4} = \frac{y-7}{5} = \frac{z-3}{7}$ has direction vector $ (4, 5, 7) $. Calculate $1 \cdot 4 + (-2) \cdot 5 + 1 \cdot 7 = 4 - 10 + 7 = 1$. They are not orthogonal.
$\frac{x-2}{3} = \frac{y-3}{3} = \frac{z-4}{4}$ has direction vector $ (3, 3, 4) $. Calculate $1 \cdot 3 + (-2) \cdot 3 + 1 \cdot 4 = 3 - 6 + 4 = 1$. They are not orthogonal.
$\frac{x-4}{3} = \frac{y-5}{4} = \frac{z-6}{3}$ has direction vector $ (3, 4, 3) $. Calculate $1 \cdot 3 + (-2) \cdot 4 + 1 \cdot 3 = 3 - 8 + 3 = -2$. They are not orthogonal.

The correct line parallel to the plane is $\frac{x-3}{4} = \frac{y-4}{5} = \frac{z-3}{6}$.

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Approach Solution -2

The equation of the plane is given by $ x - 2y + z = 0 $. The normal vector to this plane is $ \langle 1, -2, 1 \rangle $. A line is parallel to the plane if its direction vector is perpendicular to the normal of the plane. The direction vector of a line in the form $ \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} $ is $ \langle a, b, c \rangle $. We need to check each option to see which direction vector is perpendicular to $ \langle 1, -2, 1 \rangle $. Perpendicular vectors satisfy $ \mathbf{v} \cdot \mathbf{n} = 0 $ where $ \mathbf{v} $ is the direction vector and $ \mathbf{n} $ is the normal vector.

The first option $ \frac{x-3}{4} = \frac{y-4}{5} = \frac{z-3}{6} $ has direction vector $ \langle 4, 5, 6 \rangle $. Calculate the dot product:

$ 1 \cdot 4 + (-2) \cdot 5 + 1 \cdot 6 = 4 - 10 + 6 = 0 $

The dot product is zero, indicating that the direction vector is perpendicular to the normal vector of the plane, hence this option is correct.

Option	Direction Vector	Dot Product
$\frac{x-3}{4} = \frac{y-4}{5} = \frac{z-3}{6}$	$\langle 4, 5, 6 \rangle$	0 (Correct)
$\frac{x-2}{4} = \frac{y-7}{5} = \frac{z-3}{7}$	$\langle 4, 5, 7 \rangle$	8 (Incorrect)
$\frac{x-2}{3} = \frac{y-3}{3} = \frac{z-4}{4}$	$\langle 3, 3, 4 \rangle$	2 (Incorrect)
$\frac{x-4}{3} = \frac{y-5}{4} = \frac{z-6}{3}$	$\langle 3, 4, 3 \rangle$	0 (Setting Errors)

The correct direction vector is from the first option: $ \langle 4, 5, 6 \rangle $, making it parallel to the plane $ x-2y+z=0 $.