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  • Need clarity, kindly explain! - If an angle between the line, and the plane, isthen a value of k is : - Three Dimensional Geometry - JEE Main

If an angle between the line, \frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}   and the plane, x-2y-kz=3  is \cos ^{-1}\left ( \frac{2\sqrt{2}}{3} \right ),  then a value of k is :   

  • Option 1)

    -\frac{3}{5}

     

     

     

  • Option 2)

    \sqrt{\frac{5}{3}}

  • Option 3)

    -\frac{5}{3}

  • Option 4)

    \sqrt{\frac{3}{5}}

Answers (1)

best_answer

 

Angle between line and Plane (Cartesian form ) -

The angle between a line and the  line

\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} plane

a_{1}x+b_{1}y+c_{1}z+d=0 is given by

\sin \Theta = \frac{a_{1}a+b_{1}b+c_{1}c}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a^{2}+b^{2}+c^{2}}}

 

cos(90-\Theta )=\frac{\vec{n}_{1}.\vec{v}}{\left | n_{,} \right |\left .| \Theta \right |}\\\\sin\Theta =\frac{(\hat{i}-2\hat{j}-k\hat{k})(2\hat{i}+\hat{j}-2\hat{k})}{\sqrt{4+1+k^{2}}\sqrt{4+4+1}}\\\\\frac{1}{3}=\frac{\left | 2-2+2k \right |}{3\sqrt{5+k^{2}}}\\\\\left | 2k \right |=\sqrt{5+k^{2}}\\\\\\\Rightarrow k= \pm \sqrt{5/3}

-

 

 


Option 1)

-\frac{3}{5}

 

 

 

Option 2)

\sqrt{\frac{5}{3}}

Option 3)

-\frac{5}{3}

Option 4)

\sqrt{\frac{3}{5}}

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