(A) 4

(B) 5

(C) 6

(D) 3

Correct Answer: 5

Explanation

Given that a, b, and c are unit vectors, so:

|a| = |b| = |c| = 1

Using the identity:

|a − b|² = |a|² + |b|² − 2a·b

Adding all three terms:

|a − b|² + |b − c|² + |c − a|² = 6 − 2(a·b + b·c + c·a)

Given this sum equals 9, so:

6 − 2(a·b + b·c + c·a) = 9

a·b + b·c + c·a = −3/2

Now consider:

|2a + kb + kc|²

Expanding:

= 4|a|² + k²|b + c|² + 4k(a·b + a·c)

Using |b + c|² = 2 + 2b·c and the value of dot products:

|2a + kb + kc|² = 4 + k²(2 + 2b·c) + 4k(a·b + a·c)

Substituting values and simplifying gives:

|2a + kb + kc|² = 9

Solving, we get:

k = 5

Hence, the positive value of k is 5.

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