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)

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

ll₁ + mm₁ + nn₁ = 0 …(1)

and, ll₂ + mm₂ + nn₂ = 0 …(2)

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

Thus, the direction cosines of the given line are proportional to

(m₁n₂ - m₂n₁), (n₁l₂ - n₂l₁), (l₁m₂ - l₂m₁)

So, its direction cosines are

where .

we know that -

(l₁² + m₁² + n₁²) (l₂² + m₂² + n₂²) - (l₁l₂ + m₁m₂ + n₁n₂)²

= (m₁n₂ - m₂n₁)² + (n₁l₂ - n₂l₁)² + (l₁m₂ - l₂m₁)² …(3)

It is given that the given lines are perpendicular to each other. Therefore,

l₁l₂ + m₁m₂ + n₁n₂ = 0

Also, we have

l₁² + m₁² + n₁² = 1

and, l₂² + m₂² + n₂² = 1

Putting these values in (3), we get -

(m₁n₂ - m₂n₁)² + (n₁l₂ - n₂l₁)² + (l₁m₂ - l₂m₁)² = 1

⇒ λ = 1

Hence, the direction cosines of the given line are (m₁n₂ - m₂n₁), (n₁l₂ - n₂l₁), (l₁m₂ - l₂m₁)