If $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are the direction cosines of two perpendicular lines, then the direction cosine of the line which is perpendicular to both the lines, will be

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

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(

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

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\frac{1}{\sqrt{l_{1}^{2} + m_{1}^{2} + n_{1}^{2}}}, \frac{1}{\sqrt{l_{2}^{2} + m_{2}^{2} + n_{2}^{2}}}, \frac{1}{\sqrt{3}}

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\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

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Solution

The correct option is A $(m_{1} n_{2} - m_{2} n_{1}), (n_{1} l_{2} - n_{2} l_{1}), (l_{1} m_{2} - l_{2} m_{1})$

Let lines are $l_{1} x + m_{1} y + n_{1} z + d = 0 . . . (i)$
and $l_{2} x + m_{2} y + n_{2} z + d = 0... (i i)$
If lx+my+nz +d =0 is perpendicular to (i) and (ii), then , $l l_{1} + m m_{1} + n n_{1} = 0, l l_{2} + m m_{2} + n n_{2} = 0$
$\Rightarrow \frac{l}{m_{1} n_{2} - m_{2} n_{1}} = \frac{m}{n_{1} l_{2} - l_{1} n_{2}} = \frac{n}{l_{1} m_{2} - l_{2} m_{1}} = d$
Therefore, direction cosines are
$(m_{1} n_{2} - m_{2} n_{1}), (n_{1} l_{2} - l_{1} n_{2}), (l_{1} m_{2} - l_{2} m_{1})$ .

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If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m₁n₂ − m₂n₁, n₁l₂ − n₂l₁, l₁m₂ − l₂m₁.