Exams Joint Entrance Examination Mathematics 3D Geometry Question 131 Question 131 of 521 Medium NUMERICAL
A line passes through A ( 4 , − 6 , − 2 ) A(4,-6,-2) A ( 4 , − 6 , − 2 ) and B ( 16 , − 2 , 4 ) B(16,-2,4) B ( 16 , − 2 , 4 ) . The point P ( a , b , c ) P(a, b, c) P ( a , b , c ) , where a , b , c a, b, c a , b , c are non-negative integers, on the line A B A B A B lies at a distance of 21 units, from the point A A A
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x − 4 12 = x + 6 4 = z + 2 6 x − 4 6 7 = y + 6 2 7 = z + 2 3 7 = 21 ( 21 × 6 7 + 4 , 2 7 × 21 − 6 , 3 7 × 21 − 2 ) = ( 22 , 0 , 7 ) = ( a , b , c ) ∴ 324 + 144 + 16 = 22 \begin{gathered} \frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6} \\ \frac{x-4}{\frac{6}{7}}=\frac{y+6}{\frac{2}{7}}=\frac{z+2}{\frac{3}{7}}=21 \\ (21\times \frac{6}{7}+4,\frac{2}{7}\times 21-6,\frac{3}{7}\times 21-2) \\ =(22,0,7)=(a,b,c) \\ \therefore \sqrt{324+144+16}=22 \end{gathered} 12 x − 4 = 4 x + 6 = 6 z + 2 7 6 x − 4 = 7 2 y + 6 = 7 3 z + 2 = 21 ( 21 × 7 6 + 4 , 7 2 × 21 − 6 , 7 3 × 21 − 2 ) = ( 22 , 0 , 7 ) = ( a , b , c ) ∴ 324 + 144 + 16 = 22 Answer the question to unlock