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Question
If AB⊥BC, then the value of k equals, where A(2k,2,3),B(k,1,5),C(3+k,2,1)
If AB⊥BC, then the value of k equals, where A(2k,2,3),B(k,1,5),C(3+k,2,1)
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Solution
The correct option is C −3
Given A(2k,2,3),B(k,1,5)andC(3+k,2,1)AB=(k−2k)^i+(1−2)^j+(5−3)^kAB=−k^i−^j+2^kBC=(3+k−k)^i+(2−1)^j+(1−5)^kBC=3^i+^j−4^kbecause AB⊥BC→AB⋅→BC=0(−k^i−^j+2^k)⋅(3^i+^j−4^k)=0−3k−1−8=03k=−9k=−3
Given
A(2k,2,3),B(k,1,5)andC(3+k,2,1)
AB=(k−2k)^i+(1−2)^j+(5−3)^k
AB=−k^i−^j+2^k
BC=(3+k−k)^i+(2−1)^j+(1−5)^k
BC=3^i+^j−4^k
because AB⊥BC
→AB⋅→BC=0
(−k^i−^j+2^k)⋅(3^i+^j−4^k)=0
−3k−1−8=0
3k=−9
k=−3
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