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Question
Let α,β & γ be distinct real numbers. The points whose position vectors are α^i+β^j+γ^k, β^i+γ^j+α^k and γ^i+α^j+β^k
Let α,β & γ be distinct real numbers. The points whose position vectors are α^i+β^j+γ^k, β^i+γ^j+α^k and γ^i+α^j+β^k
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Solution
The correct option is C Form an equilateral triangle
Let A=α^i+β^j+γ^k
B=β^i+γ^j+α^k
C=γ^i+α^j+β^k
Then →AB=(β−α)^i+(γ−β)^j+(α−γ)^k ...(i)
→BC=(γ−β)^i+(α−γ^j+(β−α)^k ,,,(ii)
→CA=(α−γ)^i+(β−α)^j+(γ−β)^k ....(iii)
Now AB=√(β−α)2+(γ−β)2+(α−γ)2
BC=√(β−α)2+(γ−β)2+(α−γ)2
CA=√(β−α)2+(γ−β)2+(α−γ)2
Hence, all the three sides are equal in magnitude.
Therefore, ABC is an equilateral triangle.
Let A=α^i+β^j+γ^k
B=β^i+γ^j+α^k
C=γ^i+α^j+β^k
Then →AB=(β−α)^i+(γ−β)^j+(α−γ)^k ...(i)
→BC=(γ−β)^i+(α−γ^j+(β−α)^k ,,,(ii)
→CA=(α−γ)^i+(β−α)^j+(γ−β)^k ....(iii)
Now AB=√(β−α)2+(γ−β)2+(α−γ)2
BC=√(β−α)2+(γ−β)2+(α−γ)2
CA=√(β−α)2+(γ−β)2+(α−γ)2
Hence, all the three sides are equal in magnitude.
Therefore, ABC is an equilateral triangle.
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