Question

`vecr = 3hati + hatj + 2hatk + l(hati - hatj + 2hatk)` and `vecr = 2hati + hatj + 56hatk + m(3hati - 5hatj + 4hatk)`

Options

• `cos^-1 (15/8)`

• `cos^-1 ((8sqrt(3))/15)`

• `cos^-1 (15/(8sqrt(3)))`

• None of these

Answer 1

`cos^-1 ((8sqrt(3))/15)`

Explanation:

Let `phi` be the angle between the given lines. The given lines are parallel to the vectors.

`vecb_1 = hati - hatj - 2hatk` and `vecb_1 = 3hati - 5hatj - 4hatk` respectively.

The angle `phi` between then is given by

`cos phi = (vecb_1 * vecb_2)/(|vecb_1||vecb_2|)`

= `((hati - hatj - 2hatk) * (3hati - 5hatj - 4hatk))/(|hati - hatj - 2hatk||hati - hatj - 2hatk|)`

= `((1)(3) + (-1)(-5) + (-2)(-4))/(sqrt(1 + 1 + 4) sqrt(9 + 25 + 16)`

= `(3 + 5 + 8)/(sqrt(6) sqrt(50)`

= `16/(sqrt(6) sqrt(50)`

= `16/(sqrt(2) sqrt(3), 5sqrt(2)`

= `(16sqrt(3))/(3 xx 10)`

= `(8sqrt(3))/15`

⇒ `phi cos^-1  ((8sqrt(3))/15)`