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Home » If and are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are
If \mathrm{l}_{1}, \mathrm{~m}_{1}, \mathrm{n}_{1} and \mathrm{l}_{2}, \mathrm{~m}_{2}, \mathrm{n}_{2} are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are \left(\mathrm{m}_{1} \mathrm{n}_{2}-\mathrm{m}_{2} \mathrm{n}_{1}\right),\left(\mathrm{n}_{1} \mathrm{l}_{2}-\mathrm{n}_{2} \mathrm{l}_{1}\right),\left(\mathrm{l}_{1} \mathrm{~m}_{2}-\mathrm{l}_{2} \mathrm{~m}_{1}\right)
CBSE | Class 12 | Maths | Miscellaneous Exercise | NCERT | Three Dimensional Geometry
If \mathrm{l}_{1}, \mathrm{~m}_{1}, \mathrm{n}_{1} and \mathrm{l}_{2}, \mathrm{~m}_{2}, \mathrm{n}_{2} are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are \left(\mathrm{m}_{1} \mathrm{n}_{2}-\mathrm{m}_{2} \mathrm{n}_{1}\right),\left(\mathrm{n}_{1} \mathrm{l}_{2}-\mathrm{n}_{2} \mathrm{l}_{1}\right),\left(\mathrm{l}_{1} \mathrm{~m}_{2}-\mathrm{l}_{2} \mathrm{~m}_{1}\right)

Solution:

Let’s consider l, m, n be the direction cosines of the line perpendicular to each of the given lines.
Therefore, ll_{1}+m m_{1}+n n_{1}=0 \ldots(1)
And \mathrm{ll}_{2}+\mathrm{mm}_{2}+\mathrm{nn}_{2}=0 \ldots(2)
On solving eq.(1) and eq.(2) by using cross – multiplication, we obtain
\frac{1}{\mathrm{~m}_{1} \mathrm{n}_{2}-\mathrm{m}_{2} \mathrm{n}_{1}}=\frac{\mathrm{m}}{\mathrm{n}_{1} 1_{2}-\mathrm{n}_{2} \mathrm{l}_{1}}=\frac{\mathrm{n}}{1_{1} \mathrm{~m}_{2}-1_{2} \mathrm{~m}_{1}}
Therefore, the direction cosines of the given line are proportional to
\left(m_{1} n_{2}-m_{2} n_{1}\right),\left(n_{1} l_{2}-n_{2} l_{1}\right),\left(l_{1} m_{2}-l_{2} m_{1}\right)
\left(m_{1} n_{2}-m_{2} n_{1}\right),\left(n_{1} l_{2}-n_{2} l_{1}\right),\left(l_{1} m_{2}-l_{2} m_{1}\right)
Therefore, its direction cosines are
\frac{\mathrm{m}_{1} \mathrm{n}_{2}-\mathrm{m}_{2} \mathrm{n}_{1}}{\lambda}, \frac{\mathrm{n}_{1} \mathrm{l}_{2}-\mathrm{n}_{2} \mathrm{l}_{1}}{\lambda}, \frac{l_{1} \mathrm{~m}_{2}-\mathrm{l}_{2} \mathrm{~m}_{1}}{\lambda}
Where,
\lambda=\sqrt{\left(m_{1} n_{2}-m_{2} n_{1}\right)^{2}+\left(n_{1} l_{2}-n_{2} l_{1}\right)^{2}+\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2}}
Where,
\lambda=\sqrt{\left(\mathrm{m}_{1} \mathrm{ll}_{2}-\mathrm{m}_{2} \mathrm{n}_{1}\right)^{2}+\left(\mathrm{n}_{1} 1_{2}-\mathrm{n}_{2} 1_{1}\right)^{2}+\left(\mathrm{l}_{1} \mathrm{~m}_{2}-1_{2} \mathrm{~m}_{1}\right)^{2}}
It is known to us that
\begin{array}{l} \left(l_{1}^{2}+m_{1}^{2}+n_{1}^{2}\right)\left(l_{2}^{2}+m_{2}^{2}+n_{2}^{2}\right)-\left(l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\right)^{2} \\ =\left(m_{1} n_{2}-m_{2} n_{1}\right)^{2}+\left(n_{1} l_{2}-n_{2} l_{1}\right)^{2}+\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2} \ldots(3) \end{array}
Given that the given lines are perpendicular to each other.
Therefore, l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=0
Also, we have
l_{1}^{2}+m_{1}^{2}+n_{1}^{2}=1
And, l_{2}^{2}+m_{2}^{2}+n_{2}^{2}=1
On substituting these values in eq. (3), we have
\begin{array}{l} \left(m_{1} n_{2}-m_{2} n_{1}\right)^{2}+\left(n_{1} l_{2}-n_{2} l_{1}\right)^{2}+\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2}=1 \\ \lambda=1 \end{array}
As a result, the direction cosines of the given line are \left(m_{1} n_{2}-m_{2} n_{1}\right),\left(n_{1} l_{2}-n_{2} l_{1}\right),\left(l_{1} m_{2}-l_{2} m_{1}\right)

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