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Question
Show that the points whose position vectors are 4→i−3→j+→k,2→i−4→j+5→k,→i−→j from a right angled triangle.
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Solution
By the law of vectors if →a+→b=→c or →a+→b=→c then the vectors form the sides of a triangle.Let the points A,B,C have position vectors−−→OA=4^i−3^j+^k,−−→OB=2^i−4^i−4^j+5^k,−−→OC=^i−^j−−→AB=−−→OB−−−→OA=2^i−4^j+5^k−4^i+3^k−^k=2^i−^j+4^k−−→BC=−−→OC−−−→OB=^i−^j−2^i+4^j−5^k=^i+3^j−5^k−−→CA=−−→OA−−−→OC=4^i−3^j+^k−^i+^j=3^i−2^j+^kNow −−→AB.−−→BC=(−2)(−1)+(−1)(3)+(4)(−5) =2−3−20=−21≠0−−→BC.−−→CA=(−1)(3)+(3)(−2)+(−5)(1) =−3−6−5=−14≠0−−→CA.−−→AB=(3)(−2)+(−2)(−1)+(1)(4) =−6+2+4=0⇒−−→CA⊥−−→AB⇒ΔABC is right angle at A
By the law of vectors if →a+→b=→c or →a+→b=→c then the vectors form the sides of a triangle.
Let the points A,B,C have position vectors
−−→OA=4^i−3^j+^k,−−→OB=2^i−4^i−4^j+5^k,
−−→OC=^i−^j
−−→AB=−−→OB−−−→OA=2^i−4^j+5^k−4^i+3^k−^k=2^i−^j+4^k
−−→BC=−−→OC−−−→OB=^i−^j−2^i+4^j−5^k=^i+3^j−5^k
−−→CA=−−→OA−−−→OC=4^i−3^j+^k−^i+^j=3^i−2^j+^k
Now −−→AB.−−→BC=(−2)(−1)+(−1)(3)+(4)(−5)
=2−3−20=−21≠0
−−→BC.−−→CA=(−1)(3)+(3)(−2)+(−5)(1)
=−3−6−5=−14≠0
−−→CA.−−→AB=(3)(−2)+(−2)(−1)+(1)(4)
=−6+2+4=0
⇒−−→CA⊥−−→AB
⇒ΔABC is right angle at A
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