00:01
Show that the dot product and cross product are distributive using the two cases.
00:16
First, when the three vectors are co -planar and second in the general case.
00:33
For this, consider two vectors and the dot product for this is a dot b, which is given by magnitude of a in magnitude of b cost 5 and the cross product is given by a cross b is equal to magnitude of a multiplied by magnitude of b sine 5 n cap where n cap is the unit vector representing the direction of a cross b now some conditions are there but for the dot product the same unit vectors that is i dot i equals to j dot j equal to k dot k is 1 and i .j equals j .j.
01:23
Equals j .k equals k .i is equal to 0.
01:27
These are for dot product.
01:31
Now for the cross product, i cross j equals j cross k equals k cross i is 1.
01:42
And if changing order of multiplication that is j cross i equals k cross j equals i cross k will be minus one and for the same i.
01:59
I.
01:59
I equals j dot j .j equals k dot k is zero for the cross product.
02:06
Now following the diagram showing the figure are two vectors b and c and its resultant is b plus c.
02:18
From the diagram the x company, the x component of the vector bc are b cos theta 1, comma, c cos theta 2 respectively and the sum of the components is magnitude of b, cos theta 1 plus magnitude of c, cost theta 2.
02:40
The vector relation of the two vectors b c is b plus c which is making the angle of theta 3 with the horizontal.
02:48
Now the x component of this resultant is magnitude of b c c c c 3 therefore this b plus c cos theta 3 is written as magnitude of b cos theta 1 plus magnitude of c cost theta 2 now multiplying the magnitude of a on the both side we can say magnitude of a into magnitude of b plus c cost theta 3 is equals to magnitude of a into magnitude of b cost theta 1 plus magnitude of a into magnitude of c cos theta 2...