Find the angle between the pair of lines ` bar r = ( 3 hati + 2 hatj - 4hatk ) + lamda ( hati + 2hatj + 2hatk ) and bar r = ( 5hati - 2 hatk ) + mu ( 3hati + 2 hatj + 6 hatk ) `

Explanation

Let ` theta ` be the angle between the given lines.
The given lines are parallel to the vectors ` bar ( b _ 1 ) = hati + 2 hatj + 2 hatk and bar ( b _ 2 ) = 3 hati + 2hatj + 6 hatk ` respectively. Therefore, the angle ` theta ` between them is given by
` cos theta = ( bar ( b_ 1 ) * bar ( b_ 2 ) ) /( |bar ( b _ 1 ) | * |bar ( b _ 2 ) | ) `
where, ` bar ( b _ 1 ) * bar ( b _ 2 ) = ( hati + 2hatj + 2hatk ) * ( 3hati + 2 hatj + 6 hatk ) `
= ` 1 ( 3 ) + 2 ( 2 ) + 2 ( 6 ) `
` = 3 = 4 + 12 = 19 `
` | bar ( b_ 1 ) | = sqrt ( 1 ^(2) + 2 ^(2) + 2 ^ 2 ) = 3 `
and ` |bar ( b _ 2 )| = sqrt ( 3 ^ 2 + 2^2 + 6 ^ 2 ) = 7 `
` therefore cos theta = ( 19 ) /( 3 xx 7 ) = ( 19 ) /( 21) `
` therefore theta = cos ^ ( - 1 ) (( 19 ) /( 21 ))`.