00:01
In this question we are asked to find the acute angle between these two vectors by using dot product and cross product.
00:08
Now first we find the acute angle between them by using dot product.
00:12
We know a vector dot product with b vector is given by mod of a vector, mod of b vector into pose theta.
00:24
Here theta is the angle between a vector and b vector.
00:27
Here a vector is equal to 2 a x plus a y plus 3a z dot b vector is equal to 2 a x minus 3 a y plus 2 a z equal to mode of a vector is equal to square root of 2 square plus 1 square plus 3 square into mode of b vector is equal to 1 square plus negative 3 square plus 2 square into cos theta now after taking dot product this will become equal to 2 minus 3 plus 6 equals to square root of 2 square plus 1 square plus 3 square is equal to 14 into square root of 1 square plus negative 3 square is equal to 14 into square root of 1 square plus negative 3 square plus 2 square is equal to square root of 14 cos theta and from this we obtained cos theta is equal to 5 over 14 and from this we obtained theta is equal to cos inverse 5 over 14 and the value of course inverse 5 or 14 is equals to 69 .0751 degree.
02:03
Now next we will find the angle between a vector and b vector by using cross product.
02:10
We know a vector cross b vector is given by a vector mode b vector mode sine theta into and cape.
02:23
Now after taking mode in both side we obtain mode a vector cross b vector is equal to mod of a vector, mode of b vector, sine theta and mode of n -kap becomes equal to 1.
02:40
And from this we obtain sine theta is equal to a vector cross b vector over a vector mode into b -vector mode...