Đáp án:
−−→BM=−−→BA+−−→AM=−−−→AB+k−−→AC=−−−→AB+k(−−→AB+−−→BC)=(k−1)−−→AB+k−−→BC−−→CD=−−→CB+−−→BA+−−→AD=−−−→BC−−−→AB+13−−→BC=−−−→AB−23−−→BCBM⊥CD⇔−−→BM.−−→CD=0⇔[(k−1)−−→AB+k−−→BC](−−−→AB−23−−→BC)=0⇔(1−k)AB2−k.−−→BC.−−→AB−23(k−1)−−→AB.−−→BC−2k3BC2=0⇔(1−k).(2a)2−k.0−23(k−1).0−2k3.(3a)2=0⇔(4−4k)a2−6ka2=0⇔(4−4k−6k)a2=0⇔(4−10k)a2=0⇔4−10k=0⇔k=25
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