Question

Find the angle between the planes whose vector equations are $\vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) = 5$ and $\vec r.\left( {\hat i - 2\hat j + 3\hat k} \right) = 7$

Answer 1

For finding the angle between the planes we can find the angle between their normal vectors. Then we can find the cosine of the angle between the vectors by dividing the dot product of the vectors by the product of the modulus of the vectors. Then we can take the inverse of the cosine function to get the required angle.

Complete step by step solution:
We have the planes $\vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) = 5$ and $\vec r.\left( {\hat i - 2\hat j + 3\hat k} \right) = 7$
We know that a general equation of a plane is given by $\vec r.\vec n = d$ where $\vec n$ is the normal vector or the vector perpendicular to the plane.
So, the normal vectors of the given planes are given by,
${\vec n_1} = 2\hat i + 2\hat j - 3\hat k$
${\vec n_2} = \hat i - 2\hat j + 3\hat k$
Now we can find the dot product of the vectors. We know that the dot product of 2 vectors is given by multiplying their respective vector components and taking their sum.
$\Rightarrow {\vec n_1}.{\vec n_2} = \left( {2\hat i + 2\hat j - 3\hat k} \right)\left( {\hat i - 2\hat j + 3\hat k} \right)$
On multiplying the respective vector components, we get,
$\Rightarrow {\vec n_1}.{\vec n_2} = \left( {2 \times 1} \right) + \left( { - 2 \times 2} \right) + \left( { - 3 \times 3} \right)$
On simplification, we get,
$\Rightarrow {\vec n_1}.{\vec n_2} = 2 - 4 - 9$
So, we have,
$\Rightarrow {\vec n_1}.{\vec n_2} = - 11$
Now we can find the modulus of the normal vectors.
Modulus of vector $a\hat i + b\hat j + c\hat k$ is $\sqrt {{a^2} + {b^2} + {c^2}}$
So, modulus of ${\vec n_1} = 2\hat i + 2\hat j - 3\hat k$ is,
$\Rightarrow \left| {{{\vec n}_1}} \right| = \sqrt {{2^2} + {2^2} + {{\left( { - 3} \right)}^2}}$
On simplifying the squares, we get,
$\Rightarrow$ $\left| {{{\vec n}_1}} \right| = \sqrt {4 + 4 + 9}$
On adding we get,
$\Rightarrow$ $\left| {{{\vec n}_1}} \right| = \sqrt {17}$
Now take the 2nd normal vector.
${\vec n_2} = \hat i - 2\hat j + 3\hat k$
So, the modulus is
$\Rightarrow$ $\left| {{{\vec n}_2}} \right| = \sqrt {{1^2} + {{\left( { - 2} \right)}^2} + {3^2}}$
On simplifying the squares, we get,
$\Rightarrow$ $\left| {{{\vec n}_2}} \right| = \sqrt {1 + 4 + 9}$
On adding we get,
$$\Rightarrow$$$\left| {{{\vec n}_2}} \right| = \sqrt {14} $We know that$\left| {\vec a.\vec b} \right| = \left| {\vec a} \right|\left| {\vec b} \right|\cos \theta $So, we have,$\left| {{{\vec n}_1}.{{\vec n}_2}} \right| = \left| {{{\vec n}_1}} \right|\left| {{{\vec n}_2}} \right|\cos \theta $On substituting the values, we get,$ \Rightarrow \left| { - 11} \right| = \sqrt {17} \times \sqrt {14} \times \cos \theta $On rearranging, we get,$ \Rightarrow \cos \theta = \dfrac{{\left| { - 11} \right|}}{{\sqrt {17} \times \sqrt {14} }}$We can take the positive value of the numerator and multiply the denominator. So, we get,$ \Rightarrow \cos \theta = \dfrac{{11}}{{\sqrt {238} }}$To obtain the required angle, we can take the inverse of the cos function.$ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{11}}{{\sqrt {238} }}} \right)$On simplification, we get,$ \Rightarrow \theta = {\cos ^{ - 1}}\left( {0.71} \right)$Hence, we have,$ \Rightarrow \theta = 44.76^\circ $. Thus, the angle between the normal vectors is$44.76^\circ $. Therefore, the angle between the planes also will be equal to$44.76^\circ $.

So, the required solution is ${44.76}^\circ$.

Note:
The equation of the plane in its vector form is given by $\vec r.\vec n = d$ , where $\vec n$ is the normal vector. We used the equation to find the magnitude of the dot product. We know that dot products of 2 vectors give a scalar value. We must take only the modulus of the dot product to use in the equation. We must take only the positive value for finding the inverse of the trigonometric function. The exact value of the angle can be found out by using a calculator or table.