Question:
If the direction cosines of a line satisfy the relations
\[
l-m+n=0
\]
and
\[
lm+mn-4nl=0,
\]
then the direction cosines of the line are:
If the direction cosines of a line satisfy the relations
\[
l-m+n=0
\]
and
\[
lm+mn-4nl=0,
\]
then the direction cosines of the line are:
Show Hint
For direction cosines, first use the given relations to find the ratio \(l:m:n\), then apply
\[
l^2+m^2+n^2=1
\]
to determine the actual values.
Updated On: Jun 18, 2026
- \[ \left( -\frac{1}{\sqrt6}, \frac{2}{\sqrt6}, \frac{1}{\sqrt6} \right) \]
- \[ \left( \frac{1}{\sqrt6}, -\frac{2}{\sqrt6}, \frac{1}{\sqrt6} \right) \]
- \[ \left( \frac{1}{\sqrt6}, \frac{2}{\sqrt6}, -\frac{1}{\sqrt6} \right) \]
- \[ \left( \frac{1}{\sqrt6}, \frac{2}{\sqrt6}, \frac{1}{\sqrt6} \right) \]
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use the first relation.
Given \[ l-m+n=0. \] Therefore, \[ m=l+n. \]
Step 2: Substitute into the second relation.
Given \[ lm+mn-4nl=0. \] Substituting \(m=l+n\), \[ l(l+n)+(l+n)n-4ln=0. \] Expanding, \[ l^2+ln+ln+n^2-4ln=0. \] \[ l^2-2ln+n^2=0. \] \[ (l-n)^2=0. \] Hence, \[ l=n. \]
Step 3: Find \(m\).
Since \[ m=l+n, \] and \(l=n\), \[ m=2l. \] Thus, \[ l:m:n=1:2:1. \]
Step 4: Use the direction cosine condition.
Direction cosines satisfy \[ l^2+m^2+n^2=1. \] Substituting \[ m=2l,\qquad n=l, \] we get \[ l^2+(2l)^2+l^2=1. \] \[ 6l^2=1. \] \[ l=\pm\frac1{\sqrt6}. \] Hence, \[ m=\pm\frac2{\sqrt6}, \qquad n=\pm\frac1{\sqrt6}. \] Taking the positive set corresponding to the given options, \[ \left( \frac1{\sqrt6}, \frac2{\sqrt6}, \frac1{\sqrt6} \right). \]
Step 5: Final conclusion.
Therefore, the direction cosines are \[ \boxed{ \left( \frac1{\sqrt6}, \frac2{\sqrt6}, \frac1{\sqrt6} \right) } \]
Given \[ l-m+n=0. \] Therefore, \[ m=l+n. \]
Step 2: Substitute into the second relation.
Given \[ lm+mn-4nl=0. \] Substituting \(m=l+n\), \[ l(l+n)+(l+n)n-4ln=0. \] Expanding, \[ l^2+ln+ln+n^2-4ln=0. \] \[ l^2-2ln+n^2=0. \] \[ (l-n)^2=0. \] Hence, \[ l=n. \]
Step 3: Find \(m\).
Since \[ m=l+n, \] and \(l=n\), \[ m=2l. \] Thus, \[ l:m:n=1:2:1. \]
Step 4: Use the direction cosine condition.
Direction cosines satisfy \[ l^2+m^2+n^2=1. \] Substituting \[ m=2l,\qquad n=l, \] we get \[ l^2+(2l)^2+l^2=1. \] \[ 6l^2=1. \] \[ l=\pm\frac1{\sqrt6}. \] Hence, \[ m=\pm\frac2{\sqrt6}, \qquad n=\pm\frac1{\sqrt6}. \] Taking the positive set corresponding to the given options, \[ \left( \frac1{\sqrt6}, \frac2{\sqrt6}, \frac1{\sqrt6} \right). \]
Step 5: Final conclusion.
Therefore, the direction cosines are \[ \boxed{ \left( \frac1{\sqrt6}, \frac2{\sqrt6}, \frac1{\sqrt6} \right) } \]
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