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Question

# If the vectors (âˆ’bc,b2+bc,c2+bc),(a2+ac,âˆ’ac,c2+ac) and (a2+ab,b2+ab,âˆ’ab) are coplanar (where none of a, b or c is zero). Then?

A
a2+b2+c2=1
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B
a+b+c=0
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C
ab+bc+ca=0
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D
a2+b2+c2=(a+b+c)2
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Solution

## The correct option is C ab+bc+ca=0 solve:- we have to Given three vectors which are coplaner so, their scaler triple product must be zero. ⇒∣∣ ∣ ∣∣−bcb2+bcc2+bca2+ac−acc2+aca2+abb+ab−ab∣∣ ∣ ∣∣=0R1→aR1R2→bR2,R3→CR3⇒1abc∣∣ ∣ ∣∣−abca(b2+bc)a(c2+bc)b(a2+ac)−abcb(c2+ac)c(a2+ab)c(b+ab)−abc∣∣ ∣ ∣∣=0⇒∣∣ ∣ ∣∣−bca(b+c)a(c+b)b(a+c)−acb(c+a)c(a+b)c(b+a)−ab∣∣ ∣ ∣∣=0R1→R1+R2+R3⇒∣∣ ∣ ∣∣ab+bc+caab+bc+caab+bc+cab(a+c)−acb(c+a)c(a+b)c(b+a)−ab∣∣ ∣ ∣∣=0⇒(ab+bc+ca)∣∣ ∣ ∣∣1.11b(a+c)−acb(c+a)c(a+b)c(a+b)−ab∣∣ ∣ ∣∣=0c1→c1−c3,c2→c2−c3⇒(ab+bc+ca)∣∣ ∣∣0.010−(ac+bc+ba)b(c+a)ac+bc+abbc+ac+ab−ab∣∣ ∣∣=0⇒(ab+bc+ca)(0+(ab+bc+ca)2)=0⇒(ab+bc+ca)3=0⇒ab+bc+ca=0

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