Exams Joint Entrance Examination Mathematics 3D Geometry Question 124 Question 124 of 521 Medium NUMERICAL
The square of the distance of the image of the point ( 6 , 1 , 5 ) (6,1,5) ( 6 , 1 , 5 ) in the line \[\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}\], from the origin is \_\_\_\_\_\_\_\_\_\_.
Answer: 62
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P A → ← = ( 3 λ − 5 ) i ^ + ( 2 λ − 1 ) j ^ + ( 4 λ − 3 ) k ^ ( 3 λ − 5 ) 3 + ( 2 λ − 1 ) 2 + ( 4 λ − 3 ) 4 = 0 ⇒ 9 λ − 15 + 4 λ − 2 + 16 λ − 12 = 0 ⇒ 29 λ = 29 ∴ λ = 1 ∴ A ( 4 , 2 , 6 ) ∴ P ′ : P ⇒ P ′ ( 2 , 3 , 7 ) ( O P ′ ) 2 = 4 + 9 + 49 = 62 \begin{gathered} \overset{\to \leftarrow }{PA}=(3\lambda -5)\hat{i}+(2\lambda -1)\hat{j}+(4\lambda -3)\hat{k} \\ (3\lambda -5)3+(2\lambda -1)2+(4\lambda -3)4=0 \\ \Rightarrow 9\lambda -15+4\lambda -2+16\lambda -12=0 \\ \Rightarrow 29\lambda =29 \\ \therefore \lambda =1 \\ \therefore \, A(4,2,6) \\ \therefore \, {P}^{'}:P \\ \Rightarrow {P}^{'}(2,3,7) \\ {(O{P}^{'})}^{2}=4+9+49 \\ =62 \end{gathered} P A →← = ( 3 λ − 5 ) i ^ + ( 2 λ − 1 ) j ^ + ( 4 λ − 3 ) k ^ ( 3 λ − 5 ) 3 + ( 2 λ − 1 ) 2 + ( 4 λ − 3 ) 4 = 0 ⇒ 9 λ − 15 + 4 λ − 2 + 16 λ − 12 = 0 ⇒ 29 λ = 29 ∴ λ = 1 ∴ A ( 4 , 2 , 6 ) ∴ P ′ : P ⇒ P ′ ( 2 , 3 , 7 ) ( O P ′ ) 2 = 4 + 9 + 49 = 62 Answer the question to unlock