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Show that the three lines with direction cosines are mutually perpendicular.

We know that

If l_{1}, m_{1}, n_{1} and l_{2}, m_{2}, n_{2} are the direction cosines of two lines; and θ is the acute angle between the two lines; then cos θ = |l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2}|

If two lines are perpendicular, then the angle between the two is θ = 90°

⇒ For perpendicular lines, | l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} | = cos 90° = 0, i.e.

| l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} | = 0

So, in order to check if the three lines are mutually perpendicular, we compute | l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} | for all the pairs of the three lines.

Now let the direction cosines of L_{1}, L_{2} and L_{3} be l_{1}, m_{1}, n_{1}; l_{2}, m_{2}, n_{2} and l_{3}, m_{3}, n_{3}.

First, consider

⇒

⇒ L_{1}⊥ L_{2} ……(i)

Next, consider

⇒

⇒

⇒ L_{2}⊥ L_{3} …(ii)

Now, consider

⇒

⇒

⇒ L_{1}⊥ L_{3} …(iii)

∴ By (i), (ii) and (iii), we have

L_{1}, L_{2} and L_{3} are mutually perpendicular.

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