Question

`vecr = 2hati - 5hatj + hatk + lambda(3hati + 2hatj + 6hatk)` and `vecr = 2hati - 5hatj + hatk + lambda(3hati + 2hatj + 6hatk)`

Options

• `cos^-1  (19/21)`

• `cos^-1  (21/19)`

• `cos^-1  (2/9)`

• `cos^-1  (9/2)`

Answer 1

`cos^-1  (19/21)`

Explanation:

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

`b_1 = 3hati + 2hatj + 6hatk` and `b_2 = hati + 2hatj + 2hatk` respectively.

∴ The angel `phi`, between them is given by `cos phi = vecb_1  vecb_2/(|vecb_1||vecb_2|)`

= `((3hati + 2hatj + 6hatk) * (hati + 2hatj + 2hatk))/(|3hati + 2hatj + 6hatk||hati + 2hatj + 2hatk|)`

= `((3)(1) + (2)(2) + (6)(2))/(sqrt(9 + 4 + 36) sqrt(1 + 4 + 4))`

= `(3 + 4 + 12)/(sqrt(49) sqrt(9))`

= `19/(17 xx 3)`

= `19/21`

`cos^-1 (19/21)`