In the following cases,determine whether the | vedclass

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(A) The direction ratios of the normal to the planes $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$ are $(A_1, B_1, C_1)$ and $(A_2

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$\cos^{-1} \left( \frac{2}{5} \right)$

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(A) The direction ratios of the normal to the planes $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$ are $(A_1, B_1, C_1)$ and $(A_2, B_2, C_2)$ respectively.The planes are parallel if $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$.The planes are perpendicular if $A_1A_2 + B_1B_2 + C_1C_2 = 0$.The angle $	heta$ between the planes is given by $\cos 	heta = \left| \frac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}} ight|$.For the given planes $7x + 5y + 6z + 30 = 0$ and $3x - y - 10z + 4 = 0$:$A_1 = 7, B_1 = 5, C_1 = 6$$A_2 = 3, B_2 = -1, C_2 = -10$Check for perpendicularity:$A_1A_2 + B_1B_2 + C_1C_2 = (7)(3) + (5)(-1) + (6)(-10) = 21 - 5 - 60 = -44 
eq 0$.So,the planes are not perpendicular.Check for parallelism:$\frac{A_1}{A_2} = \frac{7}{3}, \frac{B_1}{B_2} = \frac{5}{-1} = -5, \frac{C_1}{C_2} = \frac{6}{-10} = -0.6$.Since $\frac{7}{3} 
eq -5 
eq -0.6$,the planes are not parallel.Calculate the angle $	heta$:$\cos 	heta = \left| \frac{-44}{\sqrt{7^2 + 5^2 + 6^2} \sqrt{3^2 + (-1)^2 + (-10)^2}} ight|$$\cos 	heta = \left| \frac{-44}{\sqrt{49 + 25 + 36} \sqrt{9 + 1 + 100}} ight| = \left| \frac{-44}{\sqrt{110} \sqrt{110}} ight| = \frac{44}{110} = \frac{2}{5}$.Therefore,$	heta = \cos^{-1} \left( \frac{2}{5} ight)$.

In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angle between them: $7x + 5y + 6z + 30 = 0$ and $3x - y - 10z + 4 = 0$.

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