Let
u ^ = x i ^ + y j ^ + z k ^ \hat u = x\hat i + y\hat j + z\hat k u ^ = x i ^ + y j ^ + z k ^ be a unit vector, so
x 2 + y 2 + z 2 = 1. x^2+y^2+z^2=1. x 2 + y 2 + z 2 = 1.
Use the angle conditions via dot products.
The given vectors are:
a ⃗ = 1 2 i ^ + 1 2 k ^ , b ⃗ = 1 2 j ^ + 1 2 k ^ , c ⃗ = 1 2 i ^ + 1 2 j ^ . \vec a=\frac{1}{\sqrt2}\hat i+\frac{1}{\sqrt2}\hat k,
\quad
\vec b=\frac{1}{\sqrt2}\hat j+\frac{1}{\sqrt2}\hat k,
\quad
\vec c=\frac{1}{\sqrt2}\hat i+\frac{1}{\sqrt2}\hat j. a = 2 1 i ^ + 2 1 k ^ , b = 2 1 j ^ + 2 1 k ^ , c = 2 1 i ^ + 2 1 j ^ . Each of these is a unit vector.
Since the angle between u ^ \hat u u ^ a ⃗ \vec a a π 2 \frac{\pi}{2} 2 π
u ^ ⋅ a ⃗ = cos π 2 = 0. \hat u\cdot \vec a = \cos\frac{\pi}{2}=0. u ^ ⋅ a = cos 2 π = 0. So,
x + z 2 = 0 ⟹ x + z = 0. (1) \frac{x+z}{\sqrt2}=0 \implies x+z=0. \tag{1} 2 x + z = 0 ⟹ x + z = 0. ( 1 ) Since the angle between u ^ \hat u u ^ b ⃗ \vec b b π 3 \frac{\pi}{3} 3 π
u ^ ⋅ b ⃗ = cos π 3 = 1 2 . \hat u\cdot \vec b = \cos\frac{\pi}{3}=\frac12. u ^ ⋅ b = cos 3 π = 2 1 . Thus,
y + z 2 = 1 2 ⟹ y + z = 1 2 . (2) \frac{y+z}{\sqrt2}=\frac12
\implies y+z=\frac{1}{\sqrt2}. \tag{2} 2 y + z = 2 1 ⟹ y + z = 2 1 . ( 2 ) Since the angle between u ^ \hat u u ^ c ⃗ \vec c c 2 π 3 \frac{2\pi}{3} 3 2 π
u ^ ⋅ c ⃗ = cos 2 π 3 = − 1 2 . \hat u\cdot \vec c = \cos\frac{2\pi}{3}=-\frac12. u ^ ⋅ c = cos 3 2 π = − 2 1 . Hence,
x + y 2 = − 1 2 ⟹ x + y = − 1 2 . (3) \frac{x+y}{\sqrt2}=-\frac12
\implies x+y=-\frac{1}{\sqrt2}. \tag{3} 2 x + y = − 2 1 ⟹ x + y = − 2 1 . ( 3 )
Solve equations (1), (2), (3).
From (1):
z = − x . z=-x. z = − x . From (3):
y = − 1 2 − x . y=-\frac{1}{\sqrt2}-x. y = − 2 1 − x . Substitute into (2):
( − 1 2 − x ) + ( − x ) = 1 2 \left(-\frac{1}{\sqrt2}-x\right)+(-x)=\frac{1}{\sqrt2} ( − 2 1 − x ) + ( − x ) = 2 1 − 1 2 − 2 x = 1 2 -\frac{1}{\sqrt2}-2x=\frac{1}{\sqrt2} − 2 1 − 2 x = 2 1 − 2 x = 2 ⟹ x = − 1 2 . -2x=\sqrt2
\implies x=-\frac{1}{\sqrt2}. − 2 x = 2 ⟹ x = − 2 1 . Then
z = − x = 1 2 , z=-x=\frac{1}{\sqrt2}, z = − x = 2 1 , and
y = − 1 2 − ( − 1 2 ) = 0. y=-\frac{1}{\sqrt2}-\left(-\frac{1}{\sqrt2}\right)=0. y = − 2 1 − ( − 2 1 ) = 0. So,
u ^ = − 1 2 i ^ + 0 j ^ + 1 2 k ^ . \hat u=-\frac{1}{\sqrt2}\hat i+0\hat j+\frac{1}{\sqrt2}\hat k. u ^ = − 2 1 i ^ + 0 j ^ + 2 1 k ^ . Check:
x 2 + y 2 + z 2 = 1 2 + 0 + 1 2 = 1 , x^2+y^2+z^2=\frac12+0+\frac12=1, x 2 + y 2 + z 2 = 2 1 + 0 + 2 1 = 1 , so it is indeed a unit vector.
Now compute
v ⃗ = 1 2 i ^ + 1 2 j ^ + 1 2 k ^ . \vec v=\frac{1}{\sqrt2}\hat i+\frac{1}{\sqrt2}\hat j+\frac{1}{\sqrt2}\hat k. v = 2 1 i ^ + 2 1 j ^ + 2 1 k ^ . Then
u ^ − v ⃗ = ( − 1 2 − 1 2 ) i ^ + ( 0 − 1 2 ) j ^ + ( 1 2 − 1 2 ) k ^ . \hat u-\vec v=
\left(-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\right)\hat i+
\left(0-\frac{1}{\sqrt2}\right)\hat j+
\left(\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\right)\hat k. u ^ − v = ( − 2 1 − 2 1 ) i ^ + ( 0 − 2 1 ) j ^ + ( 2 1 − 2 1 ) k ^ . So,
u ^ − v ⃗ = − 2 i ^ − 1 2 j ^ . \hat u-\vec v=-\sqrt2\,\hat i-\frac{1}{\sqrt2}\,\hat j. u ^ − v = − 2 i ^ − 2 1 j ^ . Therefore,
∣ u ^ − v ⃗ ∣ 2 = ( 2 ) 2 + ( 1 2 ) 2 = 2 + 1 2 = 5 2 . |\hat u-\vec v|^2 = (\sqrt2)^2+\left(\frac{1}{\sqrt2}\right)^2 = 2+\frac12=\frac52. ∣ u ^ − v ∣ 2 = ( 2 ) 2 + ( 2 1 ) 2 = 2 + 2 1 = 2 5 .
Hence the correct option is
5 2 . \boxed{\frac52}. 2 5 .