Question

find the angle between planes 3x-2y+6z-5=0 and 2x-y-2z-7=0

Answer 1

Answer:

The angle is 100.98°

Step-by-step explanation:

Given two planes 3x-2y+6z-5=0 and 2x-y-2z-7=0. we have to find the angle between the above two planes.

To find the angle between two intersecting planes we have to find the angle between normal vectors of these two planes.

For plane: 3x-2y+6z-5=0

$n_0=<3,-2,6>$ implies the vector $\vec A=3\hat{i}-2\hat{j}+6\hat{k}$

For plane: 2x-y-2z-7=0

$n_1=<2,-1,-2>$ implies the vector $\vec B=2\hat{i}-1\hat{j}-2\hat{k}$

As, $cos(\theta)=\frac{A.B}{|A||B|}$ so $\theta=cos^{-1}\frac{A.B}{|A||B|}$

$\vec A.\vec B=(3\hat{i}-2\hat{j}+6\hat{k}).(2\hat{i}-1\hat{j}-2\hat{k})=6+2-12=-4$

$|A||B|=\sqrt{3^2+2^2+6^2}\sqrt{2^2+1^2+2^2}=\sqrt{49}\sqrt{9}=7.3=21$

$\theta=cos^{-1}\frac{\vec A.\vec B}{|A||B|}=cos^{-1}({\frac{-4}{21})=100.98^{\circ}$

The angle is 100.98°