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Answer please! - Three Dimensional Geometry - JEE Main-6

If the line,\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4} meets the plane,x+2y+3z=15 at a point P, then the distance of P from the origin is :

  • Option 1)

       \sqrt{5}/2          

  • Option 2)

    2\sqrt{5}

  • Option 3)

    9/2

  • Option 4)

    7/2

 
Answers (1)
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Given Line \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}

   any  point on line is 

P\left ( 2\lambda +1,\; 3\lambda -1,\; 4\lambda +2 \right )

This point will lie on plane   x+2y+3z=15

\left ( 2\lambda +1 \right )+2\left ( 3\lambda -1 \right )+3\left ( 4\lambda +2 \right )=15

          20\lambda =10

           \lambda =\frac{1}{2}

\therefore\; \; \; P=\left ( 2,\frac{1}{2},4 \right )

Distance from origin= \sqrt{4+\frac{1}{4}+16}=\frac{9}{2}


Option 1)

   \sqrt{5}/2          

Option 2)

2\sqrt{5}

Option 3)

9/2

Option 4)

7/2

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