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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 triangle.

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Solution

We have, −−→AB=Position vector of B - Position vector of A⇒−−→AB=(^i−3^j−5^k)−(2^i−^j+^k)=−^i−2^j−6^k−−→BC= Position vector of C - Position vector of B⇒−−→BC=(3^i−4^j−4^k)−(^i−3^j−5^k)=2^i−^j+^kand−−→CA=Position vector of A - Position vector of C−−→CA=(2^i−^j+^k)−(3^i−4^j−4^k)=−^i+3^j+5^kClearly,−−→AB+−−→BC+−−→CA=→0⇒(−^i−2^j−6^k)+(2^i−^j+^k)+(−^i+3^j+5^k)=→0So, A,B and C are the vertices of a triangle.Now,−−→BC.−−→CA=(2^i−^j+^k).(−i+3^j+5^k)⇒−2−3+5=0⇒−−→BC⊥−−→CA→⇒∠BCA=π2Hence, △ABC is a right angled triangle.−−→CB×−−→CA=∣∣ ∣ ∣∣^i^j^k−21−1−135∣∣ ∣ ∣∣=^i(5+3)+^j(10+1)+^k(−6+1)=8^i+11^j−5^k|−−→CB×−−→CA|=√(8)2+(11)2+(−5)2=√64+121+25=√210 So, Area of triangle =12|−−→CB×−−→CA|=12√210 sq. units.

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