Question:
Direction cosines of line \( \frac{1 - x}{0} = y = z \) are
Direction cosines of line \( \frac{1 - x}{0} = y = z \) are
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Always ensure the variables \( x, y, z \) have a coefficient of \( +1 \) before identifying DRs.
Recall that \( L^2 + M^2 + N^2 = 1 \) for direction cosines; use this to verify options quickly.
If denominators include zero, it indicates the line is perpendicular to that specific axis.
Recall that \( L^2 + M^2 + N^2 = 1 \) for direction cosines; use this to verify options quickly.
If denominators include zero, it indicates the line is perpendicular to that specific axis.
Updated On: Sep 10, 2026
- \( 1, 1, 1 \)
- \( 0, \frac{-1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \)
- \( 1, 0, 0 \)
- \( 0, -1, -1 \)
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The Correct Option is B
Solution and Explanation
Concept:
• The standard symmetrical form of a line is \( \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \), where \( (l, m, n) \) are the direction ratios.
• Direction cosines \( (L, M, N) \) are related to direction ratios \( (l, m, n) \) by the formula \( L = \frac{l}{\sqrt{l^2 + m^2 + n^2}} \), \( M = \frac{m}{\sqrt{l^2 + m^2 + n^2}} \), and \( N = \frac{n}{\sqrt{l^2 + m^2 + n^2}} \).
• Direction cosines are unique for a directed line but can have two sets of opposite signs for an undirected line.
Step 1: Convert the given line equation into standard form
The given equation is \( \frac{1 - x}{0} = y = z \).
We rewrite it to have positive \( x \) coefficients in the numerator:
\[ \frac{-(x - 1)}{0} = \frac{y - 0}{1} = \frac{z - 0}{1} \]
Multiplying the denominators of the first term by \( -1 \):
\[ \frac{x - 1}{0} = \frac{y - 0}{1} = \frac{z - 0}{1} \]
Step 2: Identify the direction ratios (DRs) of the line
Comparing with the standard form \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \):
The direction ratios are \( a = 0 \), \( b = 1 \), and \( c = 1 \).
Step 3: Calculate the magnitude for direction cosines
Let \( D = \sqrt{a^2 + b^2 + c^2} \):
\[ D = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} \]
Step 4: Determine the direction cosines (DCs)
The DCs are given by \( \left( \frac{a}{D}, \frac{b}{D}, \frac{c}{D} \right) \):
\[ \text{DCs} = \left( \frac{0}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \]
Since direction cosines can also be represented with reversed signs (opposite direction):
\[ \text{DCs} = \left( 0, -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right) \]
This matches option (B).
• The standard symmetrical form of a line is \( \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \), where \( (l, m, n) \) are the direction ratios.
• Direction cosines \( (L, M, N) \) are related to direction ratios \( (l, m, n) \) by the formula \( L = \frac{l}{\sqrt{l^2 + m^2 + n^2}} \), \( M = \frac{m}{\sqrt{l^2 + m^2 + n^2}} \), and \( N = \frac{n}{\sqrt{l^2 + m^2 + n^2}} \).
• Direction cosines are unique for a directed line but can have two sets of opposite signs for an undirected line.
Step 1: Convert the given line equation into standard form
The given equation is \( \frac{1 - x}{0} = y = z \).
We rewrite it to have positive \( x \) coefficients in the numerator:
\[ \frac{-(x - 1)}{0} = \frac{y - 0}{1} = \frac{z - 0}{1} \]
Multiplying the denominators of the first term by \( -1 \):
\[ \frac{x - 1}{0} = \frac{y - 0}{1} = \frac{z - 0}{1} \]
Step 2: Identify the direction ratios (DRs) of the line
Comparing with the standard form \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \):
The direction ratios are \( a = 0 \), \( b = 1 \), and \( c = 1 \).
Step 3: Calculate the magnitude for direction cosines
Let \( D = \sqrt{a^2 + b^2 + c^2} \):
\[ D = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} \]
Step 4: Determine the direction cosines (DCs)
The DCs are given by \( \left( \frac{a}{D}, \frac{b}{D}, \frac{c}{D} \right) \):
\[ \text{DCs} = \left( \frac{0}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \]
Since direction cosines can also be represented with reversed signs (opposite direction):
\[ \text{DCs} = \left( 0, -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right) \]
This matches option (B).
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