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An angle between the plane, x + y + z = 5 and the line of intersection of the planes,3 x + 4 y + z -1=0  and   5 x + 8 y + 2 z +14 =0, is :

  • Option 1)

    \sin ^{-1}\left ( \sqrt{\frac{3}{17}} \right )

     

     

     

  • Option 2)

    \cos ^{-1}\left ( \sqrt{\frac{3}{17}} \right )

  • Option 3)

    \cos ^{-1}\left ( \frac{3}{\sqrt{17}} \right )

  • Option 4)

    \sin ^{-1}\left ( \frac{3}{\sqrt{17}} \right )

 

Answers (1)

best_answer

As we learned

 

Angle between two lines (Vector form ) -

Let the two lines be \vec{r}= \vec{a}+\lambda \vec{b}\, and\, \vec{r}= \vec{a_{1}}+\lambda \vec{b_{1}} .The angle between two lines will be equal to angle between their parallel vectors \vec{b}\, and \, \vec{b_{1}} .

\cos \Theta =\frac{\vec{b}\cdot \vec{b_{1}}}{\left | \vec{b} \right |\left | \vec{b_{1}} \right |}

-

 

 

3x+4y+z=1\; \; \; \; ;\; \; 5x+8y+2z=-14

=6x+8y+2z=2

x=-16;\; 4y+z=-47;

Two points will be (5,12,1)   (15,-11,-3)

\vec{r}=15\hat{i}+12\hat{j}+\hat{k}+\lambda \left ( 0\hat{i}-\hat{j}-4\hat{k} \right )

\cos \theta =\frac{\vec{b}\cdot \vec{x}}{\left | \vec{b} \right |\left | \vec{x} \right |}=\frac{1-4}{\sqrt{51}}=\sqrt{\frac{3}{17}}=\cos ^{-1}\sqrt{\frac{3}{17}}

 


Option 1)

\sin ^{-1}\left ( \sqrt{\frac{3}{17}} \right )

 

 

 

Option 2)

\cos ^{-1}\left ( \sqrt{\frac{3}{17}} \right )

Option 3)

\cos ^{-1}\left ( \frac{3}{\sqrt{17}} \right )

Option 4)

\sin ^{-1}\left ( \frac{3}{\sqrt{17}} \right )

Posted by

Himanshu

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