Exams Joint Entrance Examination Mathematics 3D Geometry Question 22 Question 22 of 521 Medium MCQ
Let P ( α , β , γ ) P(\alpha ,\beta ,\gamma ) P ( α , β , γ ) be the point on the line x − 1 2 = y + 1 − 3 = z \frac{x-1}{2}=\frac{y+1}{-3}=z 2 x − 1 = − 3 y + 1 = z at a distance 4 14 4\sqrt{14} 4 14 from the point ( 1 , − 1 , 0 ) (1,-1,0) ( 1 , − 1 , 0 ) and nearer to the origin. Then the shortest distance, between the lines x − α 1 = y − β 2 = z − γ 3 \frac{x-\alpha }{1}=\frac{y-\beta }{2}=\frac{z-\gamma }{3} 1 x − α = 2 y − β = 3 z − γ and x + 5 2 = y − 10 1 = z − 3 1 \frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1} 2 x + 5 = 1 y − 10 = 1 z − 3 , is equal to
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x − 1 2 = y + 1 − 3 = z 1 = t \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z}{1}=t 2 x − 1 = − 3 y + 1 = 1 z = t
( 1 + 2 t , − 1 − 3 t , t ) 4 t 2 + 9 t 2 + t 2 = 16 × 14 t = ± 4 ( 9 , − 13 , 4 ) , ( − 7 , 11 , − 4 ) ( α , β , γ ) = ( − 7 , 11 , − 4 ) a ⃗ = ( − 7 , 11 , − 4 ) , c ⃗ = ( − 5 , 10 , 3 ) c ⃗ − a ⃗ = ( 2 , − 1 , 7 ) b × d = ( − 1 , 5 , − 3 ) S . D = ∣ − 2 − 5 − 21 1 + 25 + 9 ∣ = 28 35 \begin{gathered} (1+2t,-1-3t,t) \\ 4{t}^{2}+9{t}^{2}+{t}^{2}=16\times 14 \\ t=\pm 4 \\ (9,-13,4),(-7,11,-4) \\ (\alpha ,\beta ,\gamma )=(-7,11,-4) \\ \vec{a}=(-7,11,-4),\vec{c}=(-5,10,3) \\ \vec{c}-\vec{a}=(2,-1,7) \\ b\times d=(-1,5,-3) \\ S.D=|\frac{-2-5-21}{\sqrt{1+25+9}}|=\frac{28}{\sqrt{35}} \end{gathered} ( 1 + 2 t , − 1 − 3 t , t ) 4 t 2 + 9 t 2 + t 2 = 16 × 14 t = ± 4 ( 9 , − 13 , 4 ) , ( − 7 , 11 , − 4 ) ( α , β , γ ) = ( − 7 , 11 , − 4 ) a = ( − 7 , 11 , − 4 ) , c = ( − 5 , 10 , 3 ) c − a = ( 2 , − 1 , 7 ) b × d = ( − 1 , 5 , − 3 ) S . D = ∣ 1 + 25 + 9 − 2 − 5 − 21 ∣ = 35 28 Answer the question to unlock