L 1 : x − 1 3 = y − 1 − 1 = z + 1 0 ⇒ r = ( 1 , 1 , − 1 ) + s ( 3 , − 1 , 0 ) , L 2 : x − 2 2 = y 0 = z + 4 α ⇒ r = ( 2 , 0 , − 4 ) + t ( 2 , 0 , α ) . \begin{aligned}
L_1&:\ \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}\quad\Rightarrow\quad
\mathbf{r}= (1,1,-1)+s(3,-1,0),\\
L_2&:\ \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}\quad\Rightarrow\quad
\mathbf{r}= (2,0,-4)+t(2,0,\alpha).
\end{aligned} L 1 L 2 : 3 x − 1 = − 1 y − 1 = 0 z + 1 ⇒ r = ( 1 , 1 , − 1 ) + s ( 3 , − 1 , 0 ) , : 2 x − 2 = 0 y = α z + 4 ⇒ r = ( 2 , 0 , − 4 ) + t ( 2 , 0 , α ) . F o r i n t e r s e c t i o n B : ( 1 , 1 , − 1 ) + s ( 3 , − 1 , 0 ) = ( 2 , 0 , − 4 ) + t ( 2 , 0 , α ) . \mathrm{For\ intersection}\ B:\ (1,1,-1)+s(3,-1,0)=(2,0,-4)+t(2,0,\alpha). For intersection B : ( 1 , 1 , − 1 ) + s ( 3 , − 1 , 0 ) = ( 2 , 0 , − 4 ) + t ( 2 , 0 , α ) . Comparing components gives the system
{ 1 + 3 s = 2 + 2 t , 1 − s = 0 , − 1 = − 4 + t α . \begin{cases}
1+3s=2+2t,\\[4pt]
1-s=0,\\[4pt]
-1= -4 + t\alpha.
\end{cases} ⎩ ⎨ ⎧ 1 + 3 s = 2 + 2 t , 1 − s = 0 , − 1 = − 4 + t α . From the second equation s = 1 s=1 s = 1 1 + 3 ⋅ 1 = 2 + 2 t ⇒ t = 1 1+3\cdot1=2+2t\Rightarrow t=1 1 + 3 ⋅ 1 = 2 + 2 t ⇒ t = 1 − 1 = − 4 + 1 ⋅ α ⇒ α = 3. -1=-4+1\cdot\alpha\Rightarrow \alpha=3. − 1 = − 4 + 1 ⋅ α ⇒ α = 3.
Thus α = 3 \alpha=3 α = 3
B = ( 1 , 1 , − 1 ) + 1 ( 3 , − 1 , 0 ) = ( 4 , 0 , − 1 ) . B=(1,1,-1)+1(3,-1,0)=(4,0,-1). B = ( 1 , 1 , − 1 ) + 1 ( 3 , − 1 , 0 ) = ( 4 , 0 , − 1 ) . F o o t P o f t h e p e r p e n d i c u l a r f r o m A = ( 1 , 1 , − 1 ) o n L 2 : \mathrm{Foot}\ P\ \mathrm{of\ the\ perpendicular\ from}\ A=(1,1,-1)\ \mathrm{on}\ L_2: Foot P of the perpendicular from A = ( 1 , 1 , − 1 ) on L 2 : With α = 3 \alpha=3 α = 3 L 2 L_2 L 2 v = ( 2 , 0 , 3 ) v=(2,0,3) v = ( 2 , 0 , 3 ) L 2 L_2 L 2 A 2 = ( 2 , 0 , − 4 ) A_2=(2,0,-4) A 2 = ( 2 , 0 , − 4 )
t 0 = ( A − A 2 ) ⋅ v v ⋅ v = ( 1 − 2 , 1 − 0 , − 1 + 4 ) ⋅ ( 2 , 0 , 3 ) 2 2 + 0 2 + 3 2 = ( − 1 , 1 , 3 ) ⋅ ( 2 , 0 , 3 ) 13 = 14 13 = 7 13 . t_0=\frac{(A-A_2)\cdot v}{v\cdot v}
=\frac{(1-2,\,1-0,\, -1+4)\cdot(2,0,3)}{2^2+0^2+3^2}
=\frac{(-1,1,3)\cdot(2,0,3)}{13}=\frac{14}{13}=\frac{7}{13}\,. t 0 = v ⋅ v ( A − A 2 ) ⋅ v = 2 2 + 0 2 + 3 2 ( 1 − 2 , 1 − 0 , − 1 + 4 ) ⋅ ( 2 , 0 , 3 ) = 13 ( − 1 , 1 , 3 ) ⋅ ( 2 , 0 , 3 ) = 13 14 = 13 7 . Hence
P = A 2 + t 0 v = ( 2 , 0 , − 4 ) + 7 13 ( 2 , 0 , 3 ) = ( 40 13 , 0 , − 31 13 ) . P=A_2+t_0 v=\Big(2,0,-4\Big)+\frac{7}{13}\Big(2,0,3\Big)
=\Big(\tfrac{40}{13},\,0,\,-\tfrac{31}{13}\Big). P = A 2 + t 0 v = ( 2 , 0 , − 4 ) + 13 7 ( 2 , 0 , 3 ) = ( 13 40 , 0 , − 13 31 ) . Compute ( P B ) 2 (PB)^2 ( P B ) 2
P B 2 = ∥ P − B ∥ 2 = ( 40 13 − 4 ) 2 + ( 0 − 0 ) 2 + ( − 31 13 + 1 ) 2 = 36 13 . PB^2=\Big\|P-B\Big\|^2
=\Big(\tfrac{40}{13}-4\Big)^2+\Big(0-0\Big)^2+\Big(-\tfrac{31}{13}+1\Big)^2
=\frac{36}{13}. P B 2 = P − B 2 = ( 13 40 − 4 ) 2 + ( 0 − 0 ) 2 + ( − 13 31 + 1 ) 2 = 13 36 . Therefore
26 α ( P B ) 2 = 26 ⋅ 3 ⋅ 36 13 = 216. 26\alpha\,(PB)^2=26\cdot 3\cdot\frac{36}{13}=216. 26 α ( P B ) 2 = 26 ⋅ 3 ⋅ 13 36 = 216. 216 \boxed{216} 216