(x^+y^+z^)2≥0 ⇒3+2Σx^⋅y^≥0⇒2Σx^⋅y^≥−3 Now, ∣x^+y^∣2+∣y^+z^∣2+∣z^+x^∣2 =6+2∑x^.y^≥6+(−3) ⇒∣x^+y^∣2+∣y^+z^∣2+∣z^+x^∣2≥3
If x, y and z are three unit vectors in threedimensional space, then the minimum value of | x+ y|^2+| y+ z|^2+| z+ x|^2
JEE Main 2014 — Mathematics Vectors & 3D Geometry
If x^,y^ and z^ are three unit vectors in threedimensional space, then the minimum value of ∣x^+y^∣2+∣y^+z^∣2+∣z^+x^∣2
Held on 12 Apr 2014 · Verified 6 Jul 2026.
Options
- A
23
- B
3
- C
33
- D
6
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