Exams Joint Entrance Examination Mathematics 3D Geometry Question 264 Question 264 of 521 Medium NUMERICAL
Let a line having direction ratios 1 , − 4 , 2 1,-4,2 1 , − 4 , 2 intersect the lines x − 7 3 = y − 1 − 1 = z + 2 1 \dfrac{x-7}{3}=\dfrac{y-1}{-1}=\dfrac{z+2}{1} 3 x − 7 = − 1 y − 1 = 1 z + 2 and x 2 = y − 7 3 = z 1 \dfrac{x}{2}=\dfrac{y-7}{3}=\dfrac{z}{1} 2 x = 3 y − 7 = 1 z at the points A A A and B B B . Then ( A B ) 2 (AB)^2 ( A B ) 2 is equal to:
Answer: 84
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A = ( 7 + 3 t , 1 − t , − 2 + t ) A=(7+3t,\ 1-t,\ -2+t) A = ( 7 + 3 t , 1 − t , − 2 + t ) , B = ( 2 s , 7 + 3 s , s ) B=(2s,\ 7+3s,\ s) B = ( 2 s , 7 + 3 s , s ) . Requiring A B ⃗ ∥ ( 1 , − 4 , 2 ) \vec{AB}\parallel(1,-4,2) A B ∥ ( 1 , − 4 , 2 ) :
\[3s-5t=16,\qquad t=s-2.\]
Solving: s = − 3 , t = − 5 s=-3,\ t=-5 s = − 3 , t = − 5 .
\[A=(-8,6,-7),\qquad B=(-6,-2,-3).\]
\[\vec{AB}=(2,-8,4) \implies AB^2=4+64+16=84.\] Answer the question to unlock