If the points (1, 1, \lambda) and (-3, 0, 1) are equidistant from the plane, 3x+4y-12z+13=0, then \lambdasatisfies the equation :

  • Option 1)

    3x^{2}-10x+7=0

  • Option 2)

    3x^{2}+10x+7=0

  • Option 3)

    3x^{2}+10x-13=0

  • Option 4)

    3x^{2}-10x+21=0

 

Answers (1)
V Vakul

As we learnt in 

Distance of a point from plane (Cartesian form) -

The length of perpendicular from P(x_{1},y_{1},z_{1}) to the plane

ax+by+cz+d= 0 is given by  \frac{\left [ ax_{1}+by_{1} +cz_{1}+d\right ]}{\left | \sqrt{a^{2}+b^{2}+c^{2}} \right |}

 

-

 

 Distance from plane are

\left | \frac{3+4-12\lambda+13}{13} \right |=\left | \frac{-9-12+13}{13} \right |

\left | 20-12\lambda \right |=8

\left | 3\lambda-5 \right |=2

3\lambda-5=2         or         3\lambda-5=-2

\lambda=\frac{7}{3}                    or            \lambda=1

These are roots of equation

3x^{2}-10x+7=0


Option 1)

3x^{2}-10x+7=0

This option is correct

Option 2)

3x^{2}+10x+7=0

This option is incorrect

Option 3)

3x^{2}+10x-13=0

This option is incorrect

Option 4)

3x^{2}-10x+21=0

This option is incorrect

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