Q. 2

# If l_{1}, m_{1}, n_{1} and l_{2}, m_{2}, 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 (m_{1}n_{2} - m_{2}n_{1}), (n_{1}l_{2} - n_{2}l_{1}), (l_{1}m_{2} - l_{2}m_{1})

Answer :

Let l, m, n be the direction cosines of the line perpendicular to each of the given lines. Then,

ll_{1} + mm_{1} + nn_{1} = 0 …(1)

and, ll_{2} + mm_{2} + nn_{2} = 0 …(2)

On solving (1) and (2) by cross - multiplication, we get -

Thus, the direction cosines of the given line 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})

So, its direction cosines are

where .

we know that -

(l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) (l_{2}^{2} + m_{2}^{2} + n_{2}^{2}) - (l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2})^{2}

= (m_{1}n_{2} - m_{2}n_{1})^{2} + (n_{1}l_{2} - n_{2}l_{1})^{2} + (l_{1}m_{2} - l_{2}m_{1})^{2} …(3)

It is 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

Putting these values in (3), we get -

(m_{1}n_{2} - m_{2}n_{1})^{2} + (n_{1}l_{2} - n_{2}l_{1})^{2} + (l_{1}m_{2} - l_{2}m_{1})^{2} = 1

⇒ λ = 1

Hence, the direction cosines of the given line 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})

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