Question

If \[\vec a = \hat i + 2\hat j - 3\hat k\] and \[\vec b = 3\hat i - \hat j + 2\hat k\] Calculate the angles between the vector $2\vec a + \vec b$ and $\vec a + 2\vec b$.

Answer 1

Hint: In the above question we need to determine the angle between $2\vec a + \vec b$ and $\vec a + 2\vec b$ then in such case we will substitute the value of the respective vector and simplify, since we know that the we need to determine the angle between them for that we must have the information of the magnitude of these two.

Complete step-by-step answer:
The first step is to determine the value of $2\vec a + \vec b$ and $\vec a + 2\vec b$ such that the value are
\[\vec a = \hat i + 2\hat j - 3\hat k\]
And
\[\vec b = 3\hat i - \hat j + 2\hat k\]
Now substituting the above value we get
$\Rightarrow$$2\vec a + \vec b = 2(\hat i + 2\hat j - 3\hat k) + 3\hat i - \hat j + 2\hat k$
On simplifying the above we get
$\Rightarrow$$2\vec a + \vec b = 2\hat i + 4\hat j - 6\hat k + 3\hat i - \hat j + 2\hat k$
On simplifying we get
$\Rightarrow$$2\vec a + \vec b = 5\hat i + 3\hat j - 4\hat k$
Now, we will determine the value of
$\Rightarrow$$\vec a + 2\vec b = \hat i + 2\hat j - 3\hat k + 2(3\hat i - \hat j + 2\hat k)$
Simplifying the above, we get
$\Rightarrow$$\vec a + 2\vec b = \hat i + 2\hat j - 3\hat k + 2(3\hat i - \hat j + 2\hat k)$
Simplifying the above, we get
$\Rightarrow$$\vec a + 2\vec b = \hat i + 2\hat j - 3\hat k + 6\hat i - 2\hat j + 4\hat k$
Hence, we get
$\Rightarrow$$\vec a + 2\vec b = 7\hat i + \hat k$
Now, let us determine the magnitude of both , we get
$\Rightarrow$$|2\vec a + \vec b| = \sqrt {{{(5)}^2} + {{(3)}^2} + {{( - 4)}^2}} = \sqrt {25 + 9 + 16} = \sqrt {50} $
Similarly, determining the magnitude of the other we get
$\Rightarrow$\[|\vec a + 2\vec b| = \sqrt {{{(7)}^2} + {{(1)}^2}} = \sqrt {50} \]
Hence, we get the magnitude of both
Now, we need to determine the solution or b=we need to determine the angle between them then in such case the formula for the determination of angle is given by
$\left( {\vec a + 2\vec b} \right)\left( {2\vec a + \vec b} \right) = \left| {2\vec a + \vec b} \right|\left| {\vec a + 2\vec b} \right|\cos \theta $
Hence, on substituting the required value we get
$\Rightarrow$$\left( {5\hat i + 3\hat j - 4\hat k} \right)\left( {7\hat i + \hat k} \right) = \left( {\sqrt {50} } \right)\left( {\sqrt {50} } \right)\cos \theta $
On simplifying the above, we get
$\Rightarrow$$35 - 4 = 50\cos \theta $
Hence, we get
$\Rightarrow$$\dfrac{{31}}{{50}} = \cos \theta $
Now, we need to determine the angle
$\Rightarrow$${\cos ^{ - 1}}\left( {\dfrac{{31}}{{50}}} \right) = \theta $

The angle between the two vectors is ${\cos ^{ - 1}}\left( {\dfrac{{31}}{{50}}} \right) $

Note: The simplification of two vectors having the dot product then in such case the multiplication then in such case $\hat i$ with $\hat j,\hat k$ is zero and vice versa , secondly the multiplication of same kind is one, Hence we get the dot product to be equal to $31$ and we need to determine the angle that means we have to shift the cosine function on the other side that forms in the inverse form.