Direction ratio of two lines are $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ then direction ratios of the line perpendicular to both the lines are

l_{1} - l_{2}, m_{1} - m_{2}, n_{1} - n_{2}

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l_{1} + l_{2}, m_{1} + m_{2}, n_{1} + n_{2}

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m_{1} n_{2} - n_{1} m_{2}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - m_{1} l_{2}

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m_{1} n_{2} - n_{1} m_{2}, n_{1} l_{2} - n_{1} l_{1}, l_{1} m_{2} - m_{1} l_{2}

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Solution

The correct option is C $m_{1} n_{2} - n_{1} m_{2}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - m_{1} l_{2}$
Let $u_{1} = (l_{1}, m_{1}, n_{1})$ be a unit vector along one line and $u_{2} = (l_{2}, m_{2}, n_{2})$ be a unit vector along other line.
As $u_{1} \times u_{2}$ is a unit vector perpendicular to both $u_{1}$ and $u_{2}$
$u_{1} \times u_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$=^i (m_{1} n_{2} - n_{1} m_{2}) -^j (n_{2} l_{1} - n_{1} l_{2}) +^k (l_{1} m_{2} - m_{1} l_{2})$
$=^i (m_{1} n_{2} - n_{1} m_{2}) +^j (n_{1} l_{2} - n_{2} l_{1}) + k (l_{1} m_{2} - m_{1} l_{2})$
So, the required direction cosines are:
$m_{1} n_{2} - n_{1} m_{2}$ , $n_{1} l_{2} - n_{2} l_{1}$ and $l_{1} m_{2} - m_{1} l_{2}$

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Similar questions

If $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines, show that the direction consines of the line perpendicular to both of these are $m_{1} n_{2} - m_{2} n_{1}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - l_{2} m_{1}$