Set vector \ (\ boldsymbol {a}, \ boldsymbol {b} \) satisfies \ (| \ boldsymbol {a} + \ boldsymbol {b} | = 2 | \ boldsymbol {a} - \ boldsymbol {b} | \) , \ (| \ boldsymbol {a} |. 3 = \) , then \ (| \ boldsymbol {b} | \) ranges \ (\ underline {\ qquad \ qquad} \) .
Method one of the questions provided \ [\ left (\ overrightarrow { OA}, \ overrightarrow {OB} \ right) = (\ boldsymbol {a}, \ boldsymbol {b}), \] wish to order \ (\ boldsymbol {a} = (3,0) \) .
Set \ (\ left | \ overrightarrow the OB} {\ right | = B \) , denoted \ (M \) of \ (AB \) midpoint, and denoted \ (| AM | = m \) , by the title \ [2 | OM | = | \ boldsymbol {a} + \ boldsymbol {b} | = 2 | \ boldsymbol {a} - \ boldsymbol {b} | = 4m \.] so \ (| OM | = 2M \) , thereby determining the point \ (m \) locus is Apollonius circle, and the circle equation \ [(x-4) ^ 2 + y ^ 2 = 4 \.] so \ (m = | AM | \ ) ranges \ ([1,3] \) , combined long midline theorem available \ [| OA | ^ 2 + | OB | ^ 2 = 2 \ left (| OM | ^ 2 + | MA | ^ 2 \ right). \] That \ [b ^ 2 = 10m ^ 2-9. \] Thus \ (B \) ranges \ ([1,9] \) .
Method two of the questions, we may assume \ [\ boldsymbol {a} = (3,0), \ boldsymbol {b} = (x, y) \.] By the known conditions given in question $
| \ A {boldsymbol } + \ boldsymbol {b} | = 2 | \ boldsymbol {a} - \ boldsymbol {b} |, $
available \ [. (x-5)
^ 2 + y ^ 2 = 16 \] so \ (X \ ) ranges \ ([1,9] \) , so \ [\ boldsymbol {b} ^ 2 = x ^ 2 + y ^ 2 = x ^ 2 + 16- (x-5) ^ 2 = 10x -9 \]. so \ (| \ boldsymbol {b} | \) ranges \ ([1,9] \) .