Question

The direction cosines of the line given by $x = y = 1 - z$ are:

• A. $1, 1, 1$
• B. $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}$
• C. $0, 0, 1$
• D. $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$

Hint:

Always make sure the coefficient of $x, y, z$ is exactly $+1$ before reading off direction ratios. For instance, transforming $1-z$ to $z-1$ changes the ratio sign from $+1$ to $-1$!

Answer 1

The Correct Option is B

Concept: The standard symmetrical form of a line equation in 3D geometry is given by: \[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \] where $a, b, c$ are the direction ratios of the line. The corresponding direction cosines $(l, m, n)$ are found by dividing each ratio by the total magnitude $\sqrt{a^2 + b^2 + c^2}$.

Step 1: Convert the given line equation into standard symmetrical form.
The given equations are: \[ x = y = 1 - z \] We rewrite this so that the variables $x, y,$ and $z$ all have a leading coefficient of $+1$: - $x$ can be written as $\frac{x - 0}{1}$ - $y$ can be written as $\frac{y - 0}{1}$ - $1 - z$ can be written as $-(z - 1)$, which is equivalent to $\frac{z - 1}{-1}$ Equating them all in standard structure: \[ \frac{x - 0}{1} = \frac{y - 0}{1} = \frac{z - 1}{-1} \]

Step 2: Identify the direction ratios.
By comparing this with the standard symmetrical form, the direction ratios $(a, b, c)$ are the denominators: \[ a = 1, \quad b = 1, \quad c = -1 \]

Step 3: Convert the direction ratios to direction cosines.
First, find the normalizing factor $\sqrt{a^2 + b^2 + c^2}$: \[ \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \] Now, the direction cosines $(l, m, n)$ are computed as: \[ l = \frac{a}{\sqrt{a^2+b^2+c^2}} = \frac{1}{\sqrt{3}} \] \[ m = \frac{b}{\sqrt{a^2+b^2+c^2}} = \frac{1}{\sqrt{3}} \] \[ n = \frac{c}{\sqrt{a^2+b^2+c^2}} = \frac{-1}{\sqrt{3}} \] Thus, the direction cosines are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$, matching option (B).