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If the angle \Theta between the line \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}  and the plane 2x-y+\sqrt{\lambda z}+4=0 is such that \sin \Theta =\frac{1}{3}, the value of \lambda  is

  • Option 1)

    -\frac{3}{5}\;

  • Option 2)

    \; \frac{5}{3}\;

  • Option 3)

    \; \frac{-4}{3}\;

  • Option 4)

    \frac{3}{4}

 

Answers (1)

best_answer

As we learnt in 

Angle between line and Plane (vector form ) -

The angle between a line \vec{r}= \vec{a}+\lambda \vec{b}and the plane \vec{r}.\vec{n}= d is given by

\sin \Theta = \frac{\vec{b}.\vec{n}}{\left | \vec{b} \right |\left | \vec{n} \right |}

- wherein

 

 Angle between line and normal to plane= cos \left ( \frac{\pi}{2}-\theta \right )=\frac{ 2-1+2\sqrt{2}}{3\times \sqrt{5+\lambda }}

sin\theta =\frac{ 2\sqrt{\lambda }}{3 \sqrt{5+\lambda }}=\frac{1}{3}\, \, \, = > \lambda =\frac{5}{3}

 


Option 1)

-\frac{3}{5}\;


Incorrect Option

 

Option 2)

\; \frac{5}{3}\;


Correct Option

 

Option 3)

\; \frac{-4}{3}\;


Incorrect Option

 

Option 4)

\frac{3}{4}


Incorrect Option

 

Posted by

Aadil

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