Find the angle between the planes 6x-3y+7z=8 and 2x+3y-2z=5?

(a) cos^{-1}\frac{11}{\sqrt{98}}

(b) cos^{-1}\frac{11}{\sqrt{1598}}

(c) cos^{-1}\frac{13}{\sqrt{198}}

(d) cos^{-1}\frac{11}{1598}

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I want to ask this question from Three Dimensional Geometry topic in chapter Three Dimensional Geometry of Mathematics – Class 12

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Right option is (b) cos^{-1}\frac{11}{\sqrt{1598}}

Easy explanation: We know that, the angle between two planes of the form A_1 x+B_1 y+C_1 z+D_1=0 and A_2 x+B_2 y+C_2 z+D_2=0 is given by

cosθ=\left |\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}}\right |

Given that, A_1=6,B_1=-3,C_1=7 and A_2=2,B_2=3,C_2=-2

cosθ=\left |\frac{6(2)-3(3)+7(-2)}{|\sqrt{6^2+(-3)^2+7^2} \sqrt{2^2+3^2+(-2)^2}|}\right |

cosθ=|\frac{-11}{\sqrt{94}.\sqrt{17}}|

θ=cos^{-1}\frac{11}{\sqrt{1598}}.