prove that area of triangle with vertics (t, t-2 ) ( t+ 2 ) , ( t + 3 ) is dependent of t
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Please find below the solution to the asked query:
We have:P(1,3,2)Q(2,−1,1)R(−1,2,3)PQ−→=Q−P=(2−1)iˆ+(−1−3)jˆ+(1−2)kˆ=iˆ−4jˆ−kˆPR−→=R−P=(−1−1)iˆ+(2−3)jˆ+(3−2)kˆ=−2iˆ−jˆ+kˆArea of triangle = 12 ∣∣PQ−→×PR−→∣∣⇒PQ−→×PR−→=∣∣∣∣∣iˆ1−2jˆ−4−1kˆ−11∣∣∣∣∣=(−4−1)iˆ−(1−2)jˆ+(−1−8)kˆ=−5iˆ−jˆ−9kˆ⇒∣∣PQ−→×PR−→∣∣=(−5)2+(−1)2+(−9)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=25+1+81‾‾‾‾‾‾‾‾‾‾‾√=107‾‾‾‾√Area of triangle = 12 ∣∣PQ−→×PR−→∣∣=107√2 unit2
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Please find below the solution to the asked query:
We have:P(1,3,2)Q(2,−1,1)R(−1,2,3)PQ−→=Q−P=(2−1)iˆ+(−1−3)jˆ+(1−2)kˆ=iˆ−4jˆ−kˆPR−→=R−P=(−1−1)iˆ+(2−3)jˆ+(3−2)kˆ=−2iˆ−jˆ+kˆArea of triangle = 12 ∣∣PQ−→×PR−→∣∣⇒PQ−→×PR−→=∣∣∣∣∣iˆ1−2jˆ−4−1kˆ−11∣∣∣∣∣=(−4−1)iˆ−(1−2)jˆ+(−1−8)kˆ=−5iˆ−jˆ−9kˆ⇒∣∣PQ−→×PR−→∣∣=(−5)2+(−1)2+(−9)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=25+1+81‾‾‾‾‾‾‾‾‾‾‾√=107‾‾‾‾√Area of triangle = 12 ∣∣PQ−→×PR−→∣∣=107√2 unit2
Hope this information will clear your doubts about this topic.
If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
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