Yes you can:
So you have 2 vectors:
Vector3 A = new Vector3(7,5,6);
Vector3 B = new Vector3(3,18,-46);
Now lets get those 10 intermediate vectors:
Vector3 C = B-A;
Vector3 d1 = (A+((C/10)*1)).normalized*A.magnitude;
Vector3 d2 = (A+((C/10)*2)).normalized*A.magnitude;
Vector3 d3 = (A+((C/10)*3)).normalized*A.magnitude;
Vector3 d4 = (A+((C/10)*4)).normalized*A.magnitude;
Vector3 d5 = (A+((C/10)*5)).normalized*A.magnitude;
Vector3 d6 = (A+((C/10)*6)).normalized*A.magnitude;
Vector3 d7 = (A+((C/10)*7)).normalized*A.magnitude;
Vector3 d8 = (A+((C/10)*8)).normalized*A.magnitude;
Vector3 d9 = (A+((C/10)*9)).normalized*A.magnitude;
Vector3 d10= (A+((C/10)*10)).normalized*A.magnitude;
now you can rotate A incrementally using something like this in your procedure:
A=d1;
yield return new WaitForSeconds(0.02f);
A=d2;
yield return new WaitForSeconds(0.02f);
A=d3;
yield return new WaitForSeconds(0.02f);
A=d4;
yield return new WaitForSeconds(0.02f);
A=d5;
yield return new WaitForSeconds(0.02f);
A=d6;
yield return new WaitForSeconds(0.02f);
A=d7;
yield return new WaitForSeconds(0.02f);
A=d8;
yield return new WaitForSeconds(0.02f);
A=d9;
yield return new WaitForSeconds(0.02f);
A=d10;
yield return new WaitForSeconds(0.02f);