Exams Joint Entrance Examination Mathematics 3D Geometry Question 128 Question 128 of 521 Medium NUMERICAL
Let the point ( − 1 , α , β ) (-1,\alpha ,\beta ) ( − 1 , α , β ) lie on the line of the shortest distance between the lines x + 2 − 3 = y − 2 4 = z − 5 2 \frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2} − 3 x + 2 = 4 y − 2 = 2 z − 5 and x + 2 − 1 = y + 6 2 = z − 1 0 \frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0} − 1 x + 2 = 2 y + 6 = 0 z − 1 . Then ( α − β ) 2 (\alpha -\beta {)}^{2} ( α − β ) 2 is equal to \_\_\_\_\_\_\_\_\_.
Answer: 25
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A B → ← ⊥ L ⃗ 1 A B → ← ⊥ L ⃗ 2 \overset{\to \leftarrow }{AB}\perp {\vec{L}}_{1}\overset{\to \leftarrow }{AB}\perp {\vec{L}}_{2} A B →← ⊥ L 1 A B →← ⊥ L 2
A B → ← = ( − 3 m − 2 + n + 2 , 4 m + 2 − 2 n + 6 , 2 m + 5 − 1 ) = ( − 3 m + n , 4 m − 2 n + 8 , 2 m + 4 ) A B → ← ⊥ L ⃗ 1 ⇒ − 3 ( − 3 m + n ) + 4 ( 4 m − 2 n + 8 ) + 2 ( 2 m + 4 ) = 0 ( 9 m + 16 m + 4 m ) + ( − 3 n − 8 n ) + 32 + 8 = 0 ⇒ 29 m − 11 n + 40 = 0 … ( 1 ) A B → ← ⊥ L ⃗ 2 ⇒ − 1 ( − 3 m + n ) + 2 ( 4 m − 2 n + 8 ) + 0 ( 2 m + 4 ) = 0 ⇒ 3 m − n + 8 m − 4 n + 16 = 0 ⇒ 11 m − 5 n + 16 = 0 ⇒ m = − 1 , n = 1 ⇒ A ≡ ( 1 , − 2 , 3 ) , B ≡ ( − 3 , − 4 , 1 ) A B ⇒ x − 1 2 = y + 2 1 = z − 3 1 ⇒ α = − 2 , β = 3 ⇒ ( α − β ) 2 = 25 \begin{gathered} \overset{\to \leftarrow }{AB}=(-3m-2+n+2,4m+2-2n+6,2m+5-1) \\ =(-3m+n,4m-2n+8,2m+4) \\ \overset{\to \leftarrow }{AB}\perp {\vec{L}}_{1} \\ \Rightarrow -3(-3m+n)+4(4m-2n+8)+2(2m+4)=0 \\ (9m+16m+4m)+(-3n-8n)+32+8=0 \\ \Rightarrow 29m-11n+40=0\, \ldots (1) \\ \overset{\to \leftarrow }{AB}\perp {\vec{L}}_{2} \\ \Rightarrow -1(-3m+n)+2(4m-2n+8)+0(2m+4)=0 \\ \Rightarrow 3m-n+8m-4n+16=0 \\ \Rightarrow 11m-5n+16=0\Rightarrow m=-1,n=1 \\ \Rightarrow A\equiv (1,-2,3),\, B\equiv (-3,-4,1) \\ AB\Rightarrow \frac{x-1}{2}=\frac{y+2}{1}=\frac{z-3}{1} \\ \Rightarrow \alpha =-2,\, \beta =3 \\ \Rightarrow (\alpha -\beta {)}^{2}=25 \end{gathered} A B →← = ( − 3 m − 2 + n + 2 , 4 m + 2 − 2 n + 6 , 2 m + 5 − 1 ) = ( − 3 m + n , 4 m − 2 n + 8 , 2 m + 4 ) A B →← ⊥ L 1 ⇒ − 3 ( − 3 m + n ) + 4 ( 4 m − 2 n + 8 ) + 2 ( 2 m + 4 ) = 0 ( 9 m + 16 m + 4 m ) + ( − 3 n − 8 n ) + 32 + 8 = 0 ⇒ 29 m − 11 n + 40 = 0 … ( 1 ) A B →← ⊥ L 2 ⇒ − 1 ( − 3 m + n ) + 2 ( 4 m − 2 n + 8 ) + 0 ( 2 m + 4 ) = 0 ⇒ 3 m − n + 8 m − 4 n + 16 = 0 ⇒ 11 m − 5 n + 16 = 0 ⇒ m = − 1 , n = 1 ⇒ A ≡ ( 1 , − 2 , 3 ) , B ≡ ( − 3 , − 4 , 1 ) A B ⇒ 2 x − 1 = 1 y + 2 = 1 z − 3 ⇒ α = − 2 , β = 3 ⇒ ( α − β ) 2 = 25 Answer the question to unlock