Given vectors
a ⃗ = α i ^ + j ^ − k ^ = ( α , 1 , − 1 ) \vec a = \alpha \hat i + \hat j - \hat k = (\alpha,1,-1) a = α i ^ + j ^ − k ^ = ( α , 1 , − 1 ) b ⃗ = 2 i ^ + j ^ − α k ^ = ( 2 , 1 , − α ) \vec b = 2\hat i + \hat j - \alpha \hat k = (2,1,-\alpha) b = 2 i ^ + j ^ − α k ^ = ( 2 , 1 , − α )
We need the projection of a ⃗ × b ⃗ \vec a \times \vec b a × b
c ⃗ = − i ^ + 2 j ^ − 2 k ^ = ( − 1 , 2 , − 2 ) \vec c = -\hat i + 2\hat j - 2\hat k = (-1,2,-2) c = − i ^ + 2 j ^ − 2 k ^ = ( − 1 , 2 , − 2 )
to be 30 30 30
Compute a ⃗ × b ⃗ \vec a \times \vec b a × b
a ⃗ × b ⃗ = ∣ i ^ j ^ k ^ α 1 − 1 2 1 − α ∣ \vec a \times \vec b =
\begin{vmatrix}
\hat i & \hat j & \hat k \\
\alpha & 1 & -1 \\
2 & 1 & -\alpha
\end{vmatrix} a × b = i ^ α 2 j ^ 1 1 k ^ − 1 − α Expanding:
a ⃗ × b ⃗ = i ^ ( 1 ( − α ) − ( − 1 ) ( 1 ) ) − j ^ ( α ( − α ) − ( − 1 ) ( 2 ) ) + k ^ ( α ( 1 ) − 1 ⋅ 2 ) \vec a \times \vec b
= \hat i\big(1(-\alpha)-(-1)(1)\big)
- \hat j\big(\alpha(-\alpha)-(-1)(2)\big)
+ \hat k\big(\alpha(1)-1\cdot 2\big) a × b = i ^ ( 1 ( − α ) − ( − 1 ) ( 1 ) ) − j ^ ( α ( − α ) − ( − 1 ) ( 2 ) ) + k ^ ( α ( 1 ) − 1 ⋅ 2 ) = i ^ ( 1 − α ) − j ^ ( − α 2 + 2 ) + k ^ ( α − 2 ) = \hat i(1-\alpha)-\hat j(-\alpha^2+2)+\hat k(\alpha-2) = i ^ ( 1 − α ) − j ^ ( − α 2 + 2 ) + k ^ ( α − 2 ) = ( 1 − α ) i ^ + ( α 2 − 2 ) j ^ + ( α − 2 ) k ^ = (1-\alpha)\hat i +(\alpha^2-2)\hat j +(\alpha-2)\hat k = ( 1 − α ) i ^ + ( α 2 − 2 ) j ^ + ( α − 2 ) k ^ So,
a ⃗ × b ⃗ = ( 1 − α , α 2 − 2 , α − 2 ) \vec a \times \vec b = (1-\alpha,\alpha^2-2,\alpha-2) a × b = ( 1 − α , α 2 − 2 , α − 2 )
Projection on c ⃗ = ( − 1 , 2 , − 2 ) \vec c = (-1,2,-2) c = ( − 1 , 2 , − 2 )
The scalar projection of a vector v ⃗ \vec v v c ⃗ \vec c c
proj c ⃗ ( v ⃗ ) = v ⃗ ⋅ c ⃗ ∣ c ⃗ ∣ \operatorname{proj}_{\vec c}(\vec v)=\frac{\vec v\cdot \vec c}{|\vec c|} proj c ( v ) = ∣ c ∣ v ⋅ c Here,
( a ⃗ × b ⃗ ) ⋅ c ⃗ = ( 1 − α ) ( − 1 ) + ( α 2 − 2 ) ( 2 ) + ( α − 2 ) ( − 2 ) (\vec a\times \vec b)\cdot \vec c
= (1-\alpha)(-1)+ (\alpha^2-2)(2)+ (\alpha-2)(-2) ( a × b ) ⋅ c = ( 1 − α ) ( − 1 ) + ( α 2 − 2 ) ( 2 ) + ( α − 2 ) ( − 2 ) Now simplify:
= − 1 + α + 2 α 2 − 4 − 2 α + 4 = -1+\alpha +2\alpha^2-4-2\alpha+4 = − 1 + α + 2 α 2 − 4 − 2 α + 4 = 2 α 2 − α − 1 = 2\alpha^2-\alpha-1 = 2 α 2 − α − 1 Also,
∣ c ⃗ ∣ = ( − 1 ) 2 + 2 2 + ( − 2 ) 2 = 1 + 4 + 4 = 3 |\vec c|=\sqrt{(-1)^2+2^2+(-2)^2} = \sqrt{1+4+4}=3 ∣ c ∣ = ( − 1 ) 2 + 2 2 + ( − 2 ) 2 = 1 + 4 + 4 = 3 Thus projection is
2 α 2 − α − 1 3 \frac{2\alpha^2-\alpha-1}{3} 3 2 α 2 − α − 1 Given this equals 30 30 30
2 α 2 − α − 1 3 = 30 \frac{2\alpha^2-\alpha-1}{3}=30 3 2 α 2 − α − 1 = 30 2 α 2 − α − 1 = 90 2\alpha^2-\alpha-1=90 2 α 2 − α − 1 = 90 2 α 2 − α − 91 = 0 2\alpha^2-\alpha-91=0 2 α 2 − α − 91 = 0
Solve the quadratic
2 α 2 − α − 91 = 0 2\alpha^2-\alpha-91=0 2 α 2 − α − 91 = 0 Using factorization:
2 α 2 − 14 α + 13 α − 91 = 0 2\alpha^2-14\alpha+13\alpha-91=0 2 α 2 − 14 α + 13 α − 91 = 0 2 α ( α − 7 ) + 13 ( α − 7 ) = 0 2\alpha(\alpha-7)+13(\alpha-7)=0 2 α ( α − 7 ) + 13 ( α − 7 ) = 0 ( α − 7 ) ( 2 α + 13 ) = 0 (\alpha-7)(2\alpha+13)=0 ( α − 7 ) ( 2 α + 13 ) = 0 So,
α = 7 or α = − 13 2 \alpha=7 \quad \text{or} \quad \alpha=-\frac{13}{2} α = 7 or α = − 2 13 Given α > 0 \alpha>0 α > 0
α = 7 \boxed{\alpha=7} α = 7
Check options
A: 15 2 \frac{15}{2} 2 15
B: 8 8 8
C: 13 2 \frac{13}{2} 2 13
D: 7 7 7
So the correct option is D .