The angle between two vectors -3i+6k and 2i+3j+k is

What is the angle between two vectors -3i+6k and 2i+3j+k Physics class 11 mcqs  First year physics mcqs

Transcript

00:00The question is the angle between two vectors minus 3i plus 6k ends plus 2i plus 3j plus

00:12k is a 45 degree b 60 degree c 90 degree d 180 degree is to find the angle we will use

00:27the formula is a dot b is equal to a b cos theta for theta this formula will become is

00:46cos inverse a dot b upon a b so theta will be equal to cos inverse is the value of a

01:05is minus 3i plus 6k dot the value of b is plus 2i plus 3j plus k divided by the magnitude of a so

01:31the magnitude of a will be under root minus 3 square plus 6 square into the magnitude of b is

01:49is the value of x is 2 so 2 square plus the value of y is 3 so y square means 3 square

02:01plus value of z is 1 means 1 square so theta will be obtained as cos inverse is minus is for

02:20simplification we will multiply the value of i with i and the value of j with j and the value

02:26of k with k because i dot i is equal to 1 j dot j is equal to 1 and k dot k is equal to 1 and in

02:37other conditions means i dot j will be 0 or j dot k will also be 0 so we will multiply same unit

02:49vectors so minus 3i plus 2i it will be minus 6 i dot i is equal to 1 plus here the value of j is

03:08not obtained so the value of j will become 0 so 6k into k so it will be 6 into 1 divided by under

03:21root the square of minus 3 is plus 9 plus the square of plus 6 is 36 into the square of 2 is

03:344 the square of 3 is 9 the square of 1 is 1 so theta will be obtained as cos inverse minus 6

03:49plus 6 divided by under root 9 plus 36 it will be 45 into under root 14 so plus 6 will be get

04:06cancelled with minus 6 so it will be 0 so 0 upon under root 45 under root 14 will be 0 is angle is

04:21obtained as cos inverse 0 is the value of cos inverse 0 is 90 degree so the correct option

04:33will be C 90 degree