Given condition
Let the angle between the unit vectors a ^ \widehat a a b ^ \widehat b b θ ∈ ( 0 , π ) \theta \in (0,\pi) θ ∈ ( 0 , π )
We are given
∣ ( a ^ + b ^ ) + 2 ( a ^ × b ^ ) ∣ = 2. \left| (\widehat a+\widehat b) + 2(\widehat a\times \widehat b) \right| = 2. ( a + b ) + 2 ( a × b ) = 2.
Use orthogonality of dot and cross parts
Note that a ^ × b ^ \widehat a \times \widehat b a × b a ^ \widehat a a b ^ \widehat b b a ^ + b ^ \widehat a+\widehat b a + b
Therefore,
∣ ( a ^ + b ^ ) + 2 ( a ^ × b ^ ) ∣ 2 = ∣ a ^ + b ^ ∣ 2 + 4 ∣ a ^ × b ^ ∣ 2 . \left| (\widehat a+\widehat b) + 2(\widehat a\times \widehat b) \right|^2
= |\widehat a+\widehat b|^2 + 4|\widehat a\times \widehat b|^2. ( a + b ) + 2 ( a × b ) 2 = ∣ a + b ∣ 2 + 4∣ a × b ∣ 2 . Since the magnitude is 2 2 2 4 4 4
∣ a ^ + b ^ ∣ 2 + 4 ∣ a ^ × b ^ ∣ 2 = 4. |\widehat a+\widehat b|^2 + 4|\widehat a\times \widehat b|^2 = 4. ∣ a + b ∣ 2 + 4∣ a × b ∣ 2 = 4.
Evaluate each term using θ \theta θ
Because a ^ , b ^ \widehat a,\widehat b a , b
∣ a ^ + b ^ ∣ 2 = ∣ a ^ ∣ 2 + ∣ b ^ ∣ 2 + 2 a ^ ⋅ b ^ = 1 + 1 + 2 cos θ = 2 + 2 cos θ . |\widehat a+\widehat b|^2 = |\widehat a|^2+|\widehat b|^2+2\widehat a\cdot\widehat b
= 1+1+2\cos\theta = 2+2\cos\theta. ∣ a + b ∣ 2 = ∣ a ∣ 2 + ∣ b ∣ 2 + 2 a ⋅ b = 1 + 1 + 2 cos θ = 2 + 2 cos θ . Also,
∣ a ^ × b ^ ∣ = sin θ . |\widehat a\times \widehat b| = \sin\theta. ∣ a × b ∣ = sin θ . Hence,
2 + 2 cos θ + 4 sin 2 θ = 4. 2+2\cos\theta + 4\sin^2\theta = 4. 2 + 2 cos θ + 4 sin 2 θ = 4. Using sin 2 θ = 1 − cos 2 θ \sin^2\theta=1-\cos^2\theta sin 2 θ = 1 − cos 2 θ
2 + 2 cos θ + 4 ( 1 − cos 2 θ ) = 4 , 2+2\cos\theta +4(1-\cos^2\theta)=4, 2 + 2 cos θ + 4 ( 1 − cos 2 θ ) = 4 , 6 + 2 cos θ − 4 cos 2 θ = 4 , 6+2\cos\theta-4\cos^2\theta=4, 6 + 2 cos θ − 4 cos 2 θ = 4 , 1 + cos θ − 2 cos 2 θ = 0 , 1+\cos\theta-2\cos^2\theta=0, 1 + cos θ − 2 cos 2 θ = 0 , 2 cos 2 θ − cos θ − 1 = 0. 2\cos^2\theta-\cos\theta-1=0. 2 cos 2 θ − cos θ − 1 = 0. Factorizing,
( 2 cos θ + 1 ) ( cos θ − 1 ) = 0. (2\cos\theta+1)(\cos\theta-1)=0. ( 2 cos θ + 1 ) ( cos θ − 1 ) = 0. So,
cos θ = 1 or cos θ = − 1 2 . \cos\theta=1 \quad \text{or} \quad \cos\theta=-\frac12. cos θ = 1 or cos θ = − 2 1 . But θ ∈ ( 0 , π ) \theta\in(0,\pi) θ ∈ ( 0 , π ) θ ≠ 0 \theta\neq 0 θ = 0 cos θ ≠ 1 \cos\theta\neq 1 cos θ = 1
cos θ = − 1 2 ⟹ θ = 2 π 3 . \cos\theta=-\frac12 \implies \theta=\frac{2\pi}{3}. cos θ = − 2 1 ⟹ θ = 3 2 π .
Check statement (S1)
We need to test
2 ∣ a ^ × b ^ ∣ = ∣ a ^ − b ^ ∣ . 2|\widehat a\times \widehat b| = |\widehat a-\widehat b|. 2∣ a × b ∣ = ∣ a − b ∣. Now,
∣ a ^ × b ^ ∣ = sin θ = sin 2 π 3 = 3 2 . |\widehat a\times \widehat b| = \sin\theta = \sin\frac{2\pi}{3}=\frac{\sqrt3}{2}. ∣ a × b ∣ = sin θ = sin 3 2 π = 2 3 . Thus,
2 ∣ a ^ × b ^ ∣ = 2 ⋅ 3 2 = 3 . 2|\widehat a\times \widehat b| = 2\cdot \frac{\sqrt3}{2}=\sqrt3. 2∣ a × b ∣ = 2 ⋅ 2 3 = 3 . Also,
∣ a ^ − b ^ ∣ 2 = ∣ a ^ ∣ 2 + ∣ b ^ ∣ 2 − 2 a ^ ⋅ b ^ = 2 − 2 cos θ . |\widehat a-\widehat b|^2 = |\widehat a|^2+|\widehat b|^2-2\widehat a\cdot\widehat b
=2-2\cos\theta. ∣ a − b ∣ 2 = ∣ a ∣ 2 + ∣ b ∣ 2 − 2 a ⋅ b = 2 − 2 cos θ . Since cos θ = − 1 2 \cos\theta=-\frac12 cos θ = − 2 1
∣ a ^ − b ^ ∣ 2 = 2 − 2 ( − 1 2 ) = 3 , |\widehat a-\widehat b|^2 = 2-2\left(-\frac12\right)=3, ∣ a − b ∣ 2 = 2 − 2 ( − 2 1 ) = 3 , so
∣ a ^ − b ^ ∣ = 3 . |\widehat a-\widehat b|=\sqrt3. ∣ a − b ∣ = 3 . Hence,
2 ∣ a ^ × b ^ ∣ = ∣ a ^ − b ^ ∣ . 2|\widehat a\times \widehat b|=|\widehat a-\widehat b|. 2∣ a × b ∣ = ∣ a − b ∣. So (S1) is true .
Check statement (S2)
The scalar projection of a ^ \widehat a a ( a ^ + b ^ ) (\widehat a+\widehat b) ( a + b )
a ^ ⋅ ( a ^ + b ^ ) ∣ a ^ + b ^ ∣ . \frac{\widehat a\cdot(\widehat a+\widehat b)}{|\widehat a+\widehat b|}. ∣ a + b ∣ a ⋅ ( a + b ) . Now,
a ^ ⋅ ( a ^ + b ^ ) = ∣ a ^ ∣ 2 + a ^ ⋅ b ^ = 1 + cos θ = 1 − 1 2 = 1 2 . \widehat a\cdot(\widehat a+\widehat b)=|\widehat a|^2+\widehat a\cdot\widehat b=1+\cos\theta=1-\frac12=\frac12. a ⋅ ( a + b ) = ∣ a ∣ 2 + a ⋅ b = 1 + cos θ = 1 − 2 1 = 2 1 . Also,
∣ a ^ + b ^ ∣ = 2 + 2 cos θ = 2 + 2 ( − 1 2 ) = 1. |\widehat a+\widehat b|=\sqrt{2+2\cos\theta}=
\sqrt{2+2\left(-\frac12\right)}=1. ∣ a + b ∣ = 2 + 2 cos θ = 2 + 2 ( − 2 1 ) = 1. Therefore projection is
1 / 2 1 = 1 2 . \frac{1/2}{1}=\frac12. 1 1/2 = 2 1 . So (S2) is true .
Conclusion
Both statements are true, so the correct option is
C \boxed{\text{C}} C