If $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are the direction cosines of two lines, then show that the direction cosines of the line perpendicular to them are proportional to $m_{1} n_{2} - m_{2} n_{1}$ , $n_{1} l_{2} - n_{2} l_{1}$ , $l_{1} m_{2} - l_{2} m_{1}$ .

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Solution

Given:
Let $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are the direction cosines of two given lines $L_{1} a n d L_{2}$ .
Let $ˆ n_{1} a n d ˆ n_{2}$ be the unit vectors along these lines $L_{1} a n d L_{2}$ .
$\to n_{1} = l_{1} i + m_{1} j + n_{1} k a n d \to n_{2} = l_{2} i + m_{2} j + n_{2} k$
The cross-product of two vectors $ˆ n_{1} \times ˆ n_{2} = | ˆ n_{1} | \cdot | ˆ n_{2} | sin 90^{\circ}^n$
$L_{1} ⊥ L_{2}$ , the angle between them is $90^{\circ}$
$ˆ n_{1} \times ˆ n_{2} =^n$
$ˆ n_{1} \times ˆ n_{2} =^n = ∣ ∣ ∣ ∣ \begin{matrix} i & j & k l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} \end{matrix} ∣ ∣ ∣ ∣$
$^n = (m_{1} n_{1} - m_{2} n_{1}) i - (l_{1} n_{2} - l_{2} n_{1}) j + (l_{1} m_{2} - l_{2} m_{1}) k$
Here, $ \hat{n} $ is a unit vector.

Thus, the direction cosines are $(m_{1} n_{1} - m_{2} n_{1}), (l_{1} n_{2} - l_{2} n_{1}), (l_{1} m_{2} - l_{2} m_{1})$ .

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