Question:
If \((l_1,m_1,n_1)\) and \((l_2,m_2,n_2)\) are the direction cosines of two lines satisfying the relations
\[
l^2+mn-6n^2=0
\]
and
\[
2l-m+3n=0,
\]
then
\[
|l_1l_2|+|m_1m_2|=
\]
If \((l_1,m_1,n_1)\) and \((l_2,m_2,n_2)\) are the direction cosines of two lines satisfying the relations
\[
l^2+mn-6n^2=0
\]
and
\[
2l-m+3n=0,
\]
then
\[
|l_1l_2|+|m_1m_2|=
\]
Show Hint
For direction cosines, always use \(l^2+m^2+n^2=1\). If two relations are given, first reduce them to direction ratios and then normalize them.
Updated On: Jun 26, 2026
- \(\frac{16}{3\sqrt{57}}\)
- \(\frac{2\sqrt{3}}{\sqrt{19}}\)
- \(\frac{4}{3\sqrt{57}}\)
- \(\frac{19}{3\sqrt{57}}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Use the given linear relation.
We are given \[ 2l-m+3n=0 \] So, \[ m=2l+3n \]
Step 2: Substitute \(m=2l+3n\) in the quadratic relation.
The other relation is \[ l^2+mn-6n^2=0 \] Substituting \(m=2l+3n\), we get \[ l^2+(2l+3n)n-6n^2=0 \] \[ l^2+2ln+3n^2-6n^2=0 \] \[ l^2+2ln-3n^2=0 \] Factorizing, \[ (l+3n)(l-n)=0 \] Hence, \[ l=n \] or \[ l=-3n \]
Step 3: Find the first set of direction cosines.
When \[ l=n, \] then \[ m=2l+3n=2n+3n=5n \] So, the direction ratios are \[ 1:5:1 \] Since direction cosines satisfy \[ l^2+m^2+n^2=1, \] we get \[ n^2+25n^2+n^2=1 \] \[ 27n^2=1 \] Thus, one set of direction cosines is \[ (l_1,m_1,n_1)=\left(\frac{1}{\sqrt{27}},\frac{5}{\sqrt{27}},\frac{1}{\sqrt{27}}\right) \]
Step 4: Find the second set of direction cosines.
When \[ l=-3n, \] then \[ m=2l+3n=2(-3n)+3n=-3n \] So, the direction ratios are \[ -3:-3:1 \] Using \[ l^2+m^2+n^2=1, \] we get \[ 9n^2+9n^2+n^2=1 \] \[ 19n^2=1 \] Thus, the second set of direction cosines can be taken as \[ (l_2,m_2,n_2)=\left(\frac{-3}{\sqrt{19}},\frac{-3}{\sqrt{19}},\frac{1}{\sqrt{19}}\right) \]
Step 5: Calculate \(|l_1l_2|+|m_1m_2|\).
Now, \[ |l_1l_2|=\left|\frac{1}{\sqrt{27}}\cdot \frac{-3}{\sqrt{19}}\right| \] \[ |l_1l_2|=\frac{3}{\sqrt{27}\sqrt{19}} \] Also, \[ |m_1m_2|=\left|\frac{5}{\sqrt{27}}\cdot \frac{-3}{\sqrt{19}}\right| \] \[ |m_1m_2|=\frac{15}{\sqrt{27}\sqrt{19}} \] Therefore, \[ |l_1l_2|+|m_1m_2| = \frac{3}{\sqrt{27}\sqrt{19}}+\frac{15}{\sqrt{27}\sqrt{19}} \] \[ =\frac{18}{\sqrt{27}\sqrt{19}} \] Since \[ \sqrt{27}=3\sqrt{3}, \] we get \[ \frac{18}{\sqrt{27}\sqrt{19}} = \frac{18}{3\sqrt{3}\sqrt{19}} \] \[ =\frac{6}{\sqrt{57}} \] Now, \[ \frac{6}{\sqrt{57}} = \frac{2\sqrt{3}}{\sqrt{19}} \]
Step 6: Final conclusion.
Hence, \[ \boxed{\frac{2\sqrt{3}}{\sqrt{19}}} \]
We are given \[ 2l-m+3n=0 \] So, \[ m=2l+3n \]
Step 2: Substitute \(m=2l+3n\) in the quadratic relation.
The other relation is \[ l^2+mn-6n^2=0 \] Substituting \(m=2l+3n\), we get \[ l^2+(2l+3n)n-6n^2=0 \] \[ l^2+2ln+3n^2-6n^2=0 \] \[ l^2+2ln-3n^2=0 \] Factorizing, \[ (l+3n)(l-n)=0 \] Hence, \[ l=n \] or \[ l=-3n \]
Step 3: Find the first set of direction cosines.
When \[ l=n, \] then \[ m=2l+3n=2n+3n=5n \] So, the direction ratios are \[ 1:5:1 \] Since direction cosines satisfy \[ l^2+m^2+n^2=1, \] we get \[ n^2+25n^2+n^2=1 \] \[ 27n^2=1 \] Thus, one set of direction cosines is \[ (l_1,m_1,n_1)=\left(\frac{1}{\sqrt{27}},\frac{5}{\sqrt{27}},\frac{1}{\sqrt{27}}\right) \]
Step 4: Find the second set of direction cosines.
When \[ l=-3n, \] then \[ m=2l+3n=2(-3n)+3n=-3n \] So, the direction ratios are \[ -3:-3:1 \] Using \[ l^2+m^2+n^2=1, \] we get \[ 9n^2+9n^2+n^2=1 \] \[ 19n^2=1 \] Thus, the second set of direction cosines can be taken as \[ (l_2,m_2,n_2)=\left(\frac{-3}{\sqrt{19}},\frac{-3}{\sqrt{19}},\frac{1}{\sqrt{19}}\right) \]
Step 5: Calculate \(|l_1l_2|+|m_1m_2|\).
Now, \[ |l_1l_2|=\left|\frac{1}{\sqrt{27}}\cdot \frac{-3}{\sqrt{19}}\right| \] \[ |l_1l_2|=\frac{3}{\sqrt{27}\sqrt{19}} \] Also, \[ |m_1m_2|=\left|\frac{5}{\sqrt{27}}\cdot \frac{-3}{\sqrt{19}}\right| \] \[ |m_1m_2|=\frac{15}{\sqrt{27}\sqrt{19}} \] Therefore, \[ |l_1l_2|+|m_1m_2| = \frac{3}{\sqrt{27}\sqrt{19}}+\frac{15}{\sqrt{27}\sqrt{19}} \] \[ =\frac{18}{\sqrt{27}\sqrt{19}} \] Since \[ \sqrt{27}=3\sqrt{3}, \] we get \[ \frac{18}{\sqrt{27}\sqrt{19}} = \frac{18}{3\sqrt{3}\sqrt{19}} \] \[ =\frac{6}{\sqrt{57}} \] Now, \[ \frac{6}{\sqrt{57}} = \frac{2\sqrt{3}}{\sqrt{19}} \]
Step 6: Final conclusion.
Hence, \[ \boxed{\frac{2\sqrt{3}}{\sqrt{19}}} \]
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