Write the vectors in component form
O P → = ( α − 1 α , 1 , 1 ) , O Q → = ( 1 , β − 1 β , 1 ) , O R → = ( 1 , 1 , 1 2 ) \overrightarrow{OP}=\left(\frac{\alpha-1}{\alpha},\,1,\,1\right),
\qquad
\overrightarrow{OQ}=\left(1,\,\frac{\beta-1}{\beta},\,1\right),
\qquad
\overrightarrow{OR}=(1,1,\tfrac12) O P = ( α α − 1 , 1 , 1 ) , O Q = ( 1 , β β − 1 , 1 ) , O R = ( 1 , 1 , 2 1 ) Let
a = α − 1 α = 1 − 1 α , b = β − 1 β = 1 − 1 β a=\frac{\alpha-1}{\alpha}=1-\frac1\alpha,
\qquad
b=\frac{\beta-1}{\beta}=1-\frac1\beta a = α α − 1 = 1 − α 1 , b = β β − 1 = 1 − β 1 So,
O P → = ( a , 1 , 1 ) , O Q → = ( 1 , b , 1 ) \overrightarrow{OP}=(a,1,1),\qquad \overrightarrow{OQ}=(1,b,1) O P = ( a , 1 , 1 ) , O Q = ( 1 , b , 1 )
Use the scalar triple product condition
Given
( O P → × O Q → ) ⋅ O R → = 0 (\overrightarrow{OP}\times \overrightarrow{OQ})\cdot \overrightarrow{OR}=0 ( O P × O Q ) ⋅ O R = 0 This is the determinant
∣ a 1 1 1 b 1 1 1 1 2 ∣ = 0 \begin{vmatrix}
a & 1 & 1\\
1 & b & 1\\
1 & 1 & \tfrac12
\end{vmatrix}=0 a 1 1 1 b 1 1 1 2 1 = 0 Now expand:
a ∣ b 1 1 1 2 ∣ − 1 ∣ 1 1 1 1 2 ∣ + 1 ∣ 1 b 1 1 ∣ = 0 a\begin{vmatrix}b & 1\\ 1 & \tfrac12\end{vmatrix}
-1\begin{vmatrix}1 & 1\\ 1 & \tfrac12\end{vmatrix}
+1\begin{vmatrix}1 & b\\ 1 & 1\end{vmatrix}=0 a b 1 1 2 1 − 1 1 1 1 2 1 + 1 1 1 b 1 = 0 a ( b 2 − 1 ) − ( 1 2 − 1 ) + ( 1 − b ) = 0 a\left(\frac b2-1\right)-\left(\frac12-1\right)+(1-b)=0 a ( 2 b − 1 ) − ( 2 1 − 1 ) + ( 1 − b ) = 0 a ( b 2 − 1 ) + 1 2 + 1 − b = 0 a\left(\frac b2-1\right)+\frac12+1-b=0 a ( 2 b − 1 ) + 2 1 + 1 − b = 0 a ( b 2 − 1 ) + 3 2 − b = 0 a\left(\frac b2-1\right)+\frac32-b=0 a ( 2 b − 1 ) + 2 3 − b = 0 Multiply by 2 2 2
a ( b − 2 ) + 3 − 2 b = 0 a(b-2)+3-2b=0 a ( b − 2 ) + 3 − 2 b = 0 Substitute a = 1 − 1 α a=1-\frac1\alpha a = 1 − α 1 b = 1 − 1 β b=1-\frac1\beta b = 1 − β 1
First compute:
b − 2 = 1 − 1 β − 2 = − 1 − 1 β b-2=1-\frac1\beta-2=-1-\frac1\beta b − 2 = 1 − β 1 − 2 = − 1 − β 1 So
a ( b − 2 ) = ( 1 − 1 α ) ( − 1 − 1 β ) a(b-2)=\left(1-\frac1\alpha\right)\left(-1-\frac1\beta\right) a ( b − 2 ) = ( 1 − α 1 ) ( − 1 − β 1 ) But it is simpler to use the determinant directly in terms of α , β \alpha,\beta α , β
Evaluate determinant directly in terms of α , β \alpha,\beta α , β
∣ α − 1 α 1 1 1 β − 1 β 1 1 1 1 2 ∣ = 0 \begin{vmatrix}
\frac{\alpha-1}{\alpha} & 1 & 1\\
1 & \frac{\beta-1}{\beta} & 1\\
1 & 1 & \frac12
\end{vmatrix}=0 α α − 1 1 1 1 β β − 1 1 1 1 2 1 = 0 Expanding:
α − 1 α ( β − 1 2 β − 1 ) − ( 1 2 − 1 ) + ( 1 − β − 1 β ) = 0 \frac{\alpha-1}{\alpha}\left(\frac{\beta-1}{2\beta}-1\right)
-\left(\frac12-1\right)
+\left(1-\frac{\beta-1}{\beta}\right)=0 α α − 1 ( 2 β β − 1 − 1 ) − ( 2 1 − 1 ) + ( 1 − β β − 1 ) = 0 Now,
β − 1 2 β − 1 = β − 1 − 2 β 2 β = − β + 1 2 β \frac{\beta-1}{2\beta}-1=\frac{\beta-1-2\beta}{2\beta}=-\frac{\beta+1}{2\beta} 2 β β − 1 − 1 = 2 β β − 1 − 2 β = − 2 β β + 1 Also,
− ( 1 2 − 1 ) = 1 2 , 1 − β − 1 β = 1 β -\left(\frac12-1\right)=\frac12,
\qquad
1-\frac{\beta-1}{\beta}=\frac1\beta − ( 2 1 − 1 ) = 2 1 , 1 − β β − 1 = β 1 Hence,
− α − 1 α ⋅ β + 1 2 β + 1 2 + 1 β = 0 -\frac{\alpha-1}{\alpha}\cdot \frac{\beta+1}{2\beta}+\frac12+\frac1\beta=0 − α α − 1 ⋅ 2 β β + 1 + 2 1 + β 1 = 0 Multiply by 2 α β 2\alpha\beta 2 α β
− ( α − 1 ) ( β + 1 ) + α β + 2 α = 0 -(\alpha-1)(\beta+1)+\alpha\beta+2\alpha=0 − ( α − 1 ) ( β + 1 ) + α β + 2 α = 0 Expand:
− ( α β + α − β − 1 ) + α β + 2 α = 0 -(\alpha\beta+\alpha-\beta-1)+\alpha\beta+2\alpha=0 − ( α β + α − β − 1 ) + α β + 2 α = 0 − α β − α + β + 1 + α β + 2 α = 0 -\alpha\beta-\alpha+\beta+1+\alpha\beta+2\alpha=0 − α β − α + β + 1 + α β + 2 α = 0 α + β + 1 = 0 \alpha+\beta+1=0 α + β + 1 = 0 So,
α + β = − 1 \alpha+\beta=-1 α + β = − 1
Use the plane condition
The point ( α , β , 2 ) (\alpha,\beta,2) ( α , β , 2 )
3 x + 3 y − z + l = 0 3x+3y-z+l=0 3 x + 3 y − z + l = 0 Substitute:
3 α + 3 β − 2 + l = 0 3\alpha+3\beta-2+l=0 3 α + 3 β − 2 + l = 0 3 ( α + β ) − 2 + l = 0 3(\alpha+\beta)-2+l=0 3 ( α + β ) − 2 + l = 0 Using α + β = − 1 \alpha+\beta=-1 α + β = − 1
3 ( − 1 ) − 2 + l = 0 3(-1)-2+l=0 3 ( − 1 ) − 2 + l = 0 − 3 − 2 + l = 0 -3-2+l=0 − 3 − 2 + l = 0 l = 5 l=5 l = 5
Final answer
5 \boxed{5} 5 The derived answer matches the stored correct answer.