Question

Find the angle between the planes whose vector equations are $\vec r.(2\hat i+2\hat j-3\hat k)=5 \ and \vec r.(3\hat i+3\hat j+5\hat k)=3$

Answer 1

Answer:

Angle between the given two planes is

$\theta=cos^{-1}[\frac{-3}{\sqrt{17}\sqrt{43}}]$

Step-by-step explanation:

Formula used:

The angle between the planes

$\vec{r}.\vec{n_1}=q_1$    and

$\vec{r}.\vec{n_2}=q_2$

is

$cos\theta=\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}$

Given Planes are

$\vec{r}.(2\vec{i}+2\vec{j}-3\vec{k})=5\\\\\vec{r}.(3\vec{i}+3\vec{j}+5\vec{k})=3$

comparing with

$\vec{r}.\vec{n_1}=q_1\:and\:\vec{r}.\vec{n_2}=q_2$

we get

$\vec{n_1}=2\vec{i}+2\vec{j}-3\vec{k}\\\\\vec{n_2}=3\vec{i}+3\vec{j}+5\vec{k}\\\\\vec{n_1}.\vec{n_2}=6+6-15=-3\\\\|\vec{n_1}|=\sqrt{2^2+2^2+(-3)^2}\\\\|\vec{n_1}|=\sqrt{4+4+9}=\sqrt{17}\\\\|\vec{n_2}|=\sqrt{3^2+3^2+5^2}\\\\|\vec{n_2}|=\sqrt{9+9+25}=\sqrt{43}$

Let $\theta$ be the angle between the given two planes.

Then,

$cos\theta=\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\\\\cos\theta=\frac{-3}{\sqrt{17}\sqrt{43}}\\\\\theta=cos^{-1}[\frac{-3}{\sqrt{17}\sqrt{43}}]$