The point of intersection of the line x-1-y-1-2-z-2-3 and the plane 2 x-3 y-z-0 is -MP PET 1989-

The correct option is D x/1=y 1/2=z+2/3=r, (say) So, x=r,y=2r+1,z=3r 2 ∴ 2r+3(2r+1)+(3r 2)=0⇒ r= 1/11 Hence, x= 1/11,y=9/11,z= 25/11. ...

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Standard XII

Mathematics

Image of a Point with Respect to a Plane

The point of ...

Question

The point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane 2x + 3y + z = 0 is
[MP PET 1989]

(0, 1, -2)

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(1, 2, 3)

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(-1, 9, -25)

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Solution

The correct option is D

$\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3} = r, (s a y) S o, x = r, y = 2 r + 1, z = 3 r - 2 ∴ 2 r + 3 (2 r + 1) + (3 r - 2) = 0 \Rightarrow r = \frac{- 1}{11} H e n c e, x = \frac{- 1}{11}, y = \frac{9}{11}, z = - \frac{25}{11} .$

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Similar questions

The point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane $2 x + 3 y + z = 0$ is:

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The point of intersection of the line x1=y−12=z+23 and the plane 2x + 3y + z = 0 is [MP PET 1989]

The correct option is D x1=y−12=z+23=r, (say)So, x=r,y=2r+1,z=3r−2∴ 2r+3(2r+1)+(3r−2)=0⇒r=−111Hence, x=−111,y=911,z=−2511.

The point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane 2x + 3y + z = 0 is
[MP PET 1989]

The correct option is D

$\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3} = r, (s a y) S o, x = r, y = 2 r + 1, z = 3 r - 2 ∴ 2 r + 3 (2 r + 1) + (3 r - 2) = 0 \Rightarrow r = \frac{- 1}{11} H e n c e, x = \frac{- 1}{11}, y = \frac{9}{11}, z = - \frac{25}{11} .$