Question

Angles between plane 3x-4y+5z =0 and 2x-y-2z=5

Answer 1

solution is explained in figure

Answer 2

Answer:

The angle between two planes is 19.94°.

Step-by-step explanation:

Given : The planes 3x-4y+5z =0 and 2x-y-2z=5

To find : Angles between the planes?

Solution :  

The angle between two planes is  

If planes are in form

$P_1=ax+by+cz=0\\\\P_2=dx+ey+fz=0$

Plane 1 vector are <3,-4,5>

Plane 2 vector are <2,-1,2>

The dot product of both the plane vectors

D=<3,-4,5>. <2,-1,2>=6+4+10=20

The modulus of vector 1

$M_1=\sqrt{3^2+(-4)^2+5^2}$

$M_1=\sqrt{9+16+25}$

$M_1=\sqrt{50}$

The modulus of vector 2

$M_2=\sqrt{2^2+(-1)^2+2^2}$

$M_2=\sqrt{4+1+4}$

$M_2=\sqrt{9}$

$M_2=3$

The angle formula is

$\cos \theta =\frac{D}{(M_1)(M_1)}$

$\cos \theta =\frac{20}{(\sqrt{50})(3)}$

$\cos \theta =\frac{20}{21.21}$

$\cos \theta =0.94$

$\theta =\cos^{-1}(0.94)$

$\theta =19.94^\circ$

Therefore, The angle between two planes is 19.94°.