The square of the distance of the image of the point ( 6 , 1 , 5 ) in the line 3 x − 1 = 2 y = 4 z − 2 , from the origin is _________

Let M ( 3 λ + 1 , 2 λ , 4 λ + 2 ) AM ⋅ b = 0 ⇒ 9 λ − 15 + 4 λ − 2 + 16 λ − 12 = 0 ⇒ 29 λ = 29 ⇒ λ = 1 M ( 4 , 2 , 6 ) , I = ( 2 , 3 , 7 ) Required Distance = 4 + 9 + 49 = 62 Ans. 62

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MathematicsVectorVector equation of a line, Vector equation of the angle bisectorsEasy2 minPYQ_2024

MathematicsEasynumerical

The square of the distance of the image of the point $(6, 1, 5)$ in the line $\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}$ , from the origin is _________

Answer:

62.00

Solution:

Let $M (3 λ + 1, 2 λ, 4 λ + 2)$ $AM \cdot b = 0$ $\Rightarrow 9 λ - 15 + 4 λ - 2 + 16 λ - 12 = 0$ $\Rightarrow 29 λ = 29$ $\Rightarrow λ = 1$ $M (4, 2, 6), I = (2, 3, 7)$ Required Distance $= 4 + 9 + 49 = 62$ Ans. 62

Stream:JEESubject:MathematicsTopic:VectorSubtopic:Vector equation of a line, Vector equation of the angle bisectors

⏱ 2mℹ️ Source: PYQ_2024

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