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Question
Show that the points A,B,C with position vectors 2^i−^j+^k,^i−3^j−5^k, and 3^i−4^j−4^k respectively,are the vertices of a right-angled triangle. Hence find the area of the rectangle.
Show that the points A,B,C with position vectors 2^i−^j+^k,^i−3^j−5^k, and 3^i−4^j−4^k respectively,are the vertices of a right-angled triangle. Hence find the area of the rectangle.
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Solution
Here −−→OA=2^i−^j+^k,−−→OB=^i−3^j−5^k,−−→OC=3^i−4^j−4^k
Now −−→AB=−−→OB−−−→OA=−^i−2^j−6^k,−−→BC=−−→OC−−−→OB=2^i−^j+^k and
−−→CA=−−→OA−−−→OC=−^i+3^j+5^k.
As −−→AB+−−→BC+−−→CA=→0,so A,B, and C form of a triangle.
Note that −−→CB.−−→CA=(−2^i+^j−^k).(−^i+3^j+5^k)=2+3−5=0 so,−−→CB⊥−−→CA
Hence ABC is a right angled triangle such that ∠C=900.
Also,−−→CB×−−→CA=∣∣
∣
∣∣^i^j^k−21−1−135∣∣
∣
∣∣=8^i+11^j−5^k
So,ar(ABC)=12∣∣−−→CB×−−→CA∣∣=12√64+121+25=√2102 sq.units.
Here −−→OA=2^i−^j+^k,−−→OB=^i−3^j−5^k,−−→OC=3^i−4^j−4^k
Now −−→AB=−−→OB−−−→OA=−^i−2^j−6^k,−−→BC=−−→OC−−−→OB=2^i−^j+^k and
−−→CA=−−→OA−−−→OC=−^i+3^j+5^k.
As −−→AB+−−→BC+−−→CA=→0,so A,B, and C form of a triangle.
Note that −−→CB.−−→CA=(−2^i+^j−^k).(−^i+3^j+5^k)=2+3−5=0 so,−−→CB⊥−−→CA
Hence ABC is a right angled triangle such that ∠C=900.
Also,−−→CB×−−→CA=∣∣ ∣ ∣∣^i^j^k−21−1−135∣∣ ∣ ∣∣=8^i+11^j−5^k
So,ar(ABC)=12∣∣−−→CB×−−→CA∣∣=12√64+121+25=√2102 sq.units.
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