Advertisements
Advertisements
Question
Find the angle between the following pair of lines:- `(x - 2)/ = (y - 1)/5 = (z + 3)/(-3)` and `(x + 2)/(-1) = (y - 4)/8 = (z - 5)/4`
Options
`sin^-1 (9sqrt(38))`
`cos^-1 (26/(9 sqrt(38)))`
`cos^-1 ((9sqrt(38))/26)`
`sin^-1 (26/(9 sqrt(34)))`
Solution
`cos^-1 (26/(9 sqrt(38)))`
Explanation:
Let `vecb_1` and `vecb_2` be the vectors parallel to,
`(x - 2)/2 = (y - 1)/5 = (z + 3)/(-3)` and `(x + 2)/(-1) = (y - 4)/8 = (z - 5)/7`
∴ `vecb_1 = 2hati + 5hatj - 3hatk, vecb_2 = - hati + 8hatj - 4hatk`
`cos = (vecb_1 * vecb_2)/(|vecb_1||vecb_2|)`
= `((2hati + 5hatj - 3hatk) * (-hati + 8hatj + 4hatk))/(|2hati + 5hatj phi - 3hatk||-hati + 8hatj + 4hatk|)`
= `(2 xx (-1) + 5 xx 8 + (-3) xx 4)/(sqrt(2^2 + 5^2 + (-3)^2 sqrt((-1)^2 + 8^2 + 4^2)`
= `26/(sqrt(38) sqrt(81)`
= `26/(9sqrt(38)`
= `cos^-1 (26/(9sqrt(38)))`