How do you compute the dot product for #6j*4k#?
1 Answer
Jul 1, 2016
0
Explanation:
#6j=0i+6j+0k=(0, 6 0) and 4k=0i+9j+4k=(0 0 4)
The dot product (a1 a2 a3).(b1 b2 b3)=((a1)(b1)+(a2)(b2)+(a3)(b3).
Here,
=(0 6 0).(0 0 4)
=(0)(0)+((6)(0)+(0)(4)
