Which of the following is the correct formula for the angle between two planes 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?

(a) cosθ=\frac{A_1 B_1 C_1}{A_2 B_2 C_2}

(b) 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 |

(c) sinθ=\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 |

(d) cosθ=A_1 A_2+B_1 B_2+C_1 C_2

I had been asked this question in an online interview.

I would like to ask this question from Three Dimensional Geometry topic in division Three Dimensional Geometry of Mathematics – Class 12

NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options

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Right option is (b) 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 |

For explanation I would say: If the planes are in the Cartesian form i.e. 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, where A_1,B_1,C_1 \,and \,A_2,B_2,C_2 are the direction ratios of the planes, then the angle between them 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 |