Let u⇀ be a vector on rectangular coordinate system with sloping angle 600, suppose that u→-i^ is geometric mean ofu and u-2i^ where i^ is the unit vector along x-axis then the value of (2+1) u is

∵angle between u→and i^ is 600 u→ . i^ =u→i^ cos 600=u→2 given that u→, u→-i^, u→-2i^ are in G.P. u→-i^2 = u→ u→-2i^ squarring on both sidesu→2+ i^2−2 u¯.i^2=u→2 u→2+4i^2−4 u→.i^u→2+ 1−2 u→22=u→2 u→2+4x1−4u→2u→2+2u→−1=0u→ =−2±222u→=2−1

Let u⇀ be a vector on rectangular coordinate system with sloping angle 600, suppose that u→-i^ is geometric mean ofu and u-2i^ where i^ is the unit vector along x-axis then the value of (2+1) u is

∵angle between u→and i^ is 600 u→ . i^ =u→i^ cos 600=u→2 given that u→, u→-i^, u→-2i^ are in G.P. u→-i^2 = u→ u→-2i^ squarring on both sidesu→2+ i^2−2 u¯.i^2=u→2 u→2+4i^2−4 u→.i^u→2+ 1−2 u→22=u→2 u→2+4x1−4u→2u→2+2u→−1=0u→ =−2±222u→=2−1

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$Let \overset{⇀}{u} b e a v e c t o r o n r e c t a n g u l a r c o o r d i n a t e s y s t e m w i t h s l o p i n g a n g l e 60^{0}, s u p p o s e t h a t |\vec{u} - \hat{i}| i s g e o m e t r i c m e a n o f$
$|u| a n d |u - 2 \hat{i}| w h e r e \hat{i} i s t h e u n i t v e c t o r a l o n g x - a x i s t h e n t h e v a l u e o f (\sqrt{2} + 1) |u| i s$

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Detailed Solution

$∵angle between \vec{u} a n d \hat{i} i s 60^{0} \vec{u} . \hat{i} = |\vec{u}| |\hat{i}| c o s 60^{0} = \frac{|\vec{u}|}{2} g i v e n t h a t |\vec{u}|, |\vec{u} - \hat{i}|, |\vec{u} - 2 \hat{i}| a r e i n G . P . {|\vec{u} - \hat{i}|}^{2} = |\vec{u}| |\vec{u} - 2 \hat{i}| s q u a r r i n g o n b o t h s i d e s$

$\begin{array}{l} {[{|\vec{u}|}^{2} + {|\hat{i}|}^{2} - 2 |\bar{u} . \hat{i}|]}^{2} = {|\vec{u}|}^{2} [{|\vec{u}|}^{2} + 4 {|\hat{i}|}^{2} - 4 |\vec{u} . \hat{i}|] \\ {[{|\vec{u}|}^{2} + 1 - 2 \frac{|\vec{u}|}{2}]}^{2} = {|\vec{u}|}^{2} [{|\vec{u}|}^{2} + 4 x 1 - 4 \frac{|\vec{u}|}{2}] \\ {|\vec{u}|}^{2} + 2 |\vec{u}| - 1 = 0 \\ |\vec{u}| = \frac{- 2 \pm 2 \sqrt{2}}{2} \\ |\vec{u}| = \sqrt{2} - 1 \end{array}$