Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 2.0 Junior Find the answer

Given vectors $a = (0, 1, -1)$ and $b = (1, 1, 0)$ , and $(a + \lambda b) \perp a$ , find the value of the real number $\lambda$ . The options are:

Pick one

Solution

Given $a = (0, 1, -1)$ and $b = (1, 1, 0)$ , and since $(a + \lambda b) \perp a$ , we have $(\lambda, 1 + \lambda, -1) \cdot (0, 1, -1) = 0$ . Solving this equation, we find $\lambda = \boxed{-2}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.