Given 𝑎 = (1, 3, −3), 𝑏⃗ = (−3, 6, 12), 𝑐 = (0, 8, 9) EXPLAIN how you would determine the exact magnitude of |2𝑎+(1/3)𝑏⃗ − 𝑐 |. *a, b, and c are all vectors.

Answer 1

2a = <2,6,-6>
(1/3)b = <-1, 2, 4>
2a + 1/3 b - c = <2,6,-6> + <-1,2,4) - <0,8,9>
= < 1, 0, -11 >

|2𝑎+(1/3)𝑏⃗ − 𝑐 |
= | < 1, 0, -11 > |
= √(1 + 0 + 121)
= √122