Find the cosine of the angle between the planes −1x+3y+1z=0 and the plane 5x+5y+4z=−4

Answer:The he cosine of the angle between the planes is [tex]\frac{14}{11\sqrt{6}}[/tex].Step-by-step explanation:Using the definition of the dot product:[tex]\cos\theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}[/tex]The given planes are[tex]-1x+3y+1z=0[/tex][tex]5x+5y+4z=-4[/tex]The angle between two normal vectors of the planes is the same as one of the angles between the planes. We can find a normal vector to each of the planes by looking at the coefficients of x, y, z.[tex]\overrightarrow{n_1}=<-1,3,1>[/tex][tex]\overrightarrow{n_2}=<5,5,4>[/tex][tex]\overrightarrow{n_1}\cdot \overrightarrow{n_2}=(-1)(5)+(3)(5)+(1)(4)=14[/tex][tex]|n_1|=\sqrt{(-1)^2+(3)^2+(1)^2}=\sqrt{11}[/tex][tex]|n_2|=\sqrt{(5)^2+(5)^2+(4)^2}=\sqrt{66}[/tex]The cosine of the angle between the planes[tex]\cos\theta =\frac{\overrightarrow{n_1}\cdot \overrightarrow{n_2}}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}[/tex][tex]\cos\theta =\frac{14}{\sqrt{11}\sqrt{66}}[/tex][tex]\cos\theta =\frac{14}{11\sqrt{6}}[/tex]Therefore the cosine of the angle between the planes is [tex]\frac{14}{11\sqrt{6}}[/tex].