Đáp án:
a) ta có:
−−→AB=(m−3,3−m)−−→AC=(1,6−m)AB→=(m−3,3−m)AC→=(1,6−m)
−−→AB.−−→AC=m−3+(3−m)(6−m)=6⇔m−3+18−9m+m2=6⇔m2−8m+9=0⇔(m−1)(m−8)=0⇔[m=1m=8AB→.AC→=m−3+(3−m)(6−m)=6⇔m−3+18−9m+m2=6⇔m2−8m+9=0⇔(m−1)(m−8)=0⇔[m=1m=8
b) −−→AC=(1,6−m)−−→BC=(4−m,3)AC→=(1,6−m)BC→=(4−m,3)
ta có:
−−→AC.−−→BC=4−m+3(6−m)=−2⇔4−m+18−3m=−2⇔4m=24⇔m=
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