已知向量 $\vec{\pi}=(\lambda+1,1), \overrightarrow{\mathrm{n}}=(\lambda+2,2)$, 若 $(\vec{\pi}+\vec{n}) \perp(\vec{\pi}-\overrightarrow{\mathrm{n}})$, 则 $\lambda=(\qquad)$
Gaokao · Math · previous-year question
- A.-4
- B.-3
- C.-2
- D.-1
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