a)→b+→c=(3;0;−3)⇒→a(→b+→c)=6
|→a+→b|=|(4;−1;−1)|=3√2
b)cos(→a,→b)=→a.→b|→a|.|→b|=2√10.√6=1√15
cos(→a,→c)=→a.→c|→a|.|→c|=5√10.√6=52√15
cos(→a+→b,→c)=(→a+→b)→c|→a+→b|.|→c|=8√18.√6=43√3
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