Question

如圖,在三棱柱ABC—A′B′C′中,點(diǎn)E、F、H、K分別為AC′、CB′、A′B、B′C′ 的中點(diǎn),G為△ABC的重心.從K、H、G、B′中取一點(diǎn)作為P,使得該棱柱恰有2條棱與平面PEF平行,則P為(    )

A.K                      B.H                   C.G                  D.B′

Answer 1

答案:C

解析:當(dāng)P點(diǎn)與K點(diǎn)重合時(shí),面PEF即為面KEF,因?yàn)镵F與三棱柱三條側(cè)棱都平行,不滿足題設(shè)條件.當(dāng)P點(diǎn)與H重合時(shí),面PEF即為面HEF,而面HEF與三棱柱兩底面均平行,有六條棱平行于面HEF,不合題意.當(dāng)P點(diǎn)與B′點(diǎn)重合時(shí),面PEF即為面B′EF,此時(shí)三棱柱棱中只有一條棱AB與它平行,不合題意.當(dāng)P點(diǎn)與G點(diǎn)重合時(shí),面PEF即為面GEF,此時(shí)恰有三棱柱的兩條棱AB、A′B′與面GEF平行,滿足題意.故選項(xiàng)為C.