The value of a, for which the points A,B,C with position vectors 2^i−^j+^k,^i−3^j−5^k,a^i−3^j+^k respectively are the vertices of a right angled triangle with C=π2 are:
A
1
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B
2
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C
−1
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D
−2
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Solution
The correct options are A1 D2 −−→BC=−−→OC−−−→OB =a^i−3^j+^k−(^i−3^j−5^k) =(a−1)^i+6^k −−→CA=−−→OA−−−→OC =2^i−^j+^k−(a^i−3^j+^k) =(2−a)^i+2^j −−→BC.−−→CA=[(a−1)^i+6^k].[(2−a)^i+2^j] =(a−1)(2−a)+0 since ^i.^i=1,^i.^j=0 ⇒(a−1)(2−a)=0 ∴a=1,2