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Angle between planes 2x-3y+4z-1=0 and 7x+6y+z-5=0 is

  • Option 1)

    \frac{\pi}{6}

  • Option 2)

    \frac{\pi}{3}

  • Option 3)

    \frac{\pi}{2}

  • Option 4)

    \pi

 

Answers (1)

best_answer

As we have learned

Angle between two planes (Cartesian form) -

Let the two planes be

ax+by+cz+d=0\: and \: a_{1}x+b_{1}y+c_{1}z+d_{1}=0

then the angle between them is defined as the angle between their normals

\cos \Theta = \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}

-

 

 Here a_1=2,b_1=-3,c_1=4 and a_2=7,b_2=6,c_2=1

\therefore cos\theta= \frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{\Sigma a_1^2}\sqrt{\Sigma a_2^2}}

=\frac{14+(-3)(6)+(4)}{\sqrt{29}\sqrt{86}}=0

\Rightarrow \theta=\frac{\pi}{2}


Option 1)

\frac{\pi}{6}

Option 2)

\frac{\pi}{3}

Option 3)

\frac{\pi}{2}

Option 4)

\pi

Posted by

Himanshu

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