Since a ^ \hat a a ^ b ⃗ = i ^ − 2 j ^ + 3 k ^ , c ⃗ = 2 i ^ + 3 j ^ − k ^ , \vec b=\hat i-2\hat j+3\hat k,\qquad \vec c=2\hat i+3\hat j-\hat k, b = i ^ − 2 j ^ + 3 k ^ , c = 2 i ^ + 3 j ^ − k ^ , b ⃗ × c ⃗ \vec b\times \vec c b × c
Compute the cross product:
b ⃗ × c ⃗ = ∣ i ^ j ^ k ^ 1 − 2 3 2 3 − 1 ∣ \vec b\times \vec c=
\begin{vmatrix}
\hat i&\hat j&\hat k\\
1&-2&3\\
2&3&-1
\end{vmatrix} b × c = i ^ 1 2 j ^ − 2 3 k ^ 3 − 1 = i ^ ( ( − 2 ) ( − 1 ) − 3 ⋅ 3 ) − j ^ ( 1 ( − 1 ) − 3 ⋅ 2 ) + k ^ ( 1 ⋅ 3 − ( − 2 ) ⋅ 2 ) =\hat i\big((-2)(-1)-3\cdot 3\big)-\hat j\big(1(-1)-3\cdot 2\big)+\hat k\big(1\cdot 3-(-2)\cdot 2\big) = i ^ ( ( − 2 ) ( − 1 ) − 3 ⋅ 3 ) − j ^ ( 1 ( − 1 ) − 3 ⋅ 2 ) + k ^ ( 1 ⋅ 3 − ( − 2 ) ⋅ 2 ) = i ^ ( 2 − 9 ) − j ^ ( − 1 − 6 ) + k ^ ( 3 + 4 ) = − 7 i ^ + 7 j ^ + 7 k ^ . =\hat i(2-9)-\hat j(-1-6)+\hat k(3+4)
=-7\hat i+7\hat j+7\hat k. = i ^ ( 2 − 9 ) − j ^ ( − 1 − 6 ) + k ^ ( 3 + 4 ) = − 7 i ^ + 7 j ^ + 7 k ^ . So a unit vector perpendicular to both is
a ^ = ± − i ^ + j ^ + k ^ 3 . \hat a=\pm \frac{-\hat i+\hat j+\hat k}{\sqrt3}. a ^ = ± 3 − i ^ + j ^ + k ^ .
Use the given angle with i ^ + j ^ + k ^ \hat i+\hat j+\hat k i ^ + j ^ + k ^
Let d ⃗ = i ^ + j ^ + k ^ \vec d=\hat i+\hat j+\hat k d = i ^ + j ^ + k ^ ∣ d ⃗ ∣ = 3 . |\vec d|=\sqrt3. ∣ d ∣ = 3 .
cos θ = a ^ ⋅ d ⃗ ∣ a ^ ∣ ∣ d ⃗ ∣ = a ^ ⋅ d ⃗ 3 . \cos\theta=\frac{\hat a\cdot \vec d}{|\hat a|\,|\vec d|}=\frac{\hat a\cdot \vec d}{\sqrt3}. cos θ = ∣ a ^ ∣ ∣ d ∣ a ^ ⋅ d = 3 a ^ ⋅ d . Given
cos θ = − 1 3 , \cos\theta=-\frac13, cos θ = − 3 1 , so
a ^ ⋅ d ⃗ = − 3 3 = − 1 3 . \hat a\cdot \vec d=-\frac{\sqrt3}{3}=-\frac1{\sqrt3}. a ^ ⋅ d = − 3 3 = − 3 1 . Check the two possibilities:
For a ^ = − i ^ + j ^ + k ^ 3 \hat a=\dfrac{-\hat i+\hat j+\hat k}{\sqrt3} a ^ = 3 − i ^ + j ^ + k ^
a ^ ⋅ d ⃗ = − 1 + 1 + 1 3 = 1 3 , \hat a\cdot \vec d=\frac{-1+1+1}{\sqrt3}=\frac1{\sqrt3}, a ^ ⋅ d = 3 − 1 + 1 + 1 = 3 1 , which gives cos θ = 1 3 \cos\theta=\frac13 cos θ = 3 1
a ^ = i ^ − j ^ − k ^ 3 . \hat a=\frac{\hat i-\hat j-\hat k}{\sqrt3}. a ^ = 3 i ^ − j ^ − k ^ . Indeed,
a ^ ⋅ d ⃗ = 1 − 1 − 1 3 = − 1 3 , \hat a\cdot \vec d=\frac{1-1-1}{\sqrt3}=-\frac1{\sqrt3}, a ^ ⋅ d = 3 1 − 1 − 1 = − 3 1 , and hence cos θ = − 1 3 \cos\theta=-\frac13 cos θ = − 3 1
Now use that a ^ \hat a a ^ π 3 \frac\pi3 3 π
v ⃗ = i ^ + α j ^ + k ^ . \vec v=\hat i+\alpha\hat j+\hat k. v = i ^ + α j ^ + k ^ . Since cos π 3 = 1 2 \cos\frac\pi3=\frac12 cos 3 π = 2 1
a ^ ⋅ v ⃗ ∣ a ^ ∣ ∣ v ⃗ ∣ = 1 2 . \frac{\hat a\cdot \vec v}{|\hat a|\,|\vec v|}=\frac12. ∣ a ^ ∣ ∣ v ∣ a ^ ⋅ v = 2 1 . Now ∣ a ^ ∣ = 1 |\hat a|=1 ∣ a ^ ∣ = 1
a ^ ⋅ v ⃗ = 1 − α − 1 3 = − α 3 . \hat a\cdot \vec v=\frac{1-\alpha-1}{\sqrt3}=\frac{-\alpha}{\sqrt3}. a ^ ⋅ v = 3 1 − α − 1 = 3 − α . Also,
∣ v ⃗ ∣ = 1 + α 2 + 1 = α 2 + 2 . |\vec v|=\sqrt{1+\alpha^2+1}=\sqrt{\alpha^2+2}. ∣ v ∣ = 1 + α 2 + 1 = α 2 + 2 . Therefore,
− α / 3 α 2 + 2 = 1 2 . \frac{-\alpha/\sqrt3}{\sqrt{\alpha^2+2}}=\frac12. α 2 + 2 − α / 3 = 2 1 . So
− α 3 α 2 + 2 = 1 2 . \frac{-\alpha}{\sqrt3\sqrt{\alpha^2+2}}=\frac12. 3 α 2 + 2 − α = 2 1 . Multiply through:
− 2 α = 3 α 2 + 2 . -2\alpha=\sqrt3\sqrt{\alpha^2+2}. − 2 α = 3 α 2 + 2 . Since the right side is positive, we must have α < 0 \alpha<0 α < 0
Square both sides:
4 α 2 = 3 ( α 2 + 2 ) 4\alpha^2=3(\alpha^2+2) 4 α 2 = 3 ( α 2 + 2 ) 4 α 2 = 3 α 2 + 6 4\alpha^2=3\alpha^2+6 4 α 2 = 3 α 2 + 6 α 2 = 6. \alpha^2=6. α 2 = 6. Thus
α = ± 6 . \alpha=\pm\sqrt6. α = ± 6 . But from the sign condition α < 0 \alpha<0 α < 0
α = − 6 . \alpha=-\sqrt6. α = − 6 .
Hence the correct option is
− 6 . \boxed{-\sqrt6}. − 6 .