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Let r_vec be a vector perpendicular to (a_vec + b_vec + c_vec), where [a_vec b_vec c_vec] = 2 (scalar triple product). If r_vec = l*(b_vec x c_vec) + m*(c_vec x a_vec) + n*(a_vec x b_vec), then the value of (l + m + n) equals:

1. 0
2. 2
3. -2
4. Depends on direction of r_vec

Correct answer: 0

Solution

Condition: r_vec. (a_vec + b_vec + c_vec) = 0. Substitute r_vec = l*(b_vec x c_vec) + m*(c_vec x a_vec) + n*(a_vec x b_vec). Dot product with (a_vec + b_vec + c_vec): r_vec. a_vec + r_vec. b_vec + r_vec. c_vec = 0. Now: [l*(b_vec x c_vec) + m*(c_vec x a_vec) + n*(a_vec x b_vec)]. (a_vec + b_vec + c_vec). Each cross product dotted with each vector: (b_vec x c_vec).a_vec = [a b c] = 2; (b_vec x c_vec).b_vec = 0; (b_vec x c_vec).c_vec = 0. Similarly for others. So the dot product = l*[a b c]*1 + m*[a b c]*1 + n*[a b c]*1 =... wait, let me be precise: [(b x c).a + (b x c).b + (b x c).c]*l + [(c x a).a + (c x a).b + (c x a).c]*m + [(a x b).a + (a x b).b + (a x b).c]*n. Non-zero terms only: l*(b x c).a = l*2; m*(c x a).b = m*2; n*(a x b).c = n*2. Sum = 2l + 2m + 2n = 0 => l + m + n = 0.

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