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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...

If a⃗=i^+2j^−3k^\vec a = \hat i + 2\hat j - 3\hat k and b⃗=3i^−j^+2k^\vec b = 3\hat i - \hat j + 2\hat k Calculate the angles between the vector 2a⃗+b⃗2\vec a + \vec b and a⃗+2b⃗\vec a + 2\vec b.

Explanation

Solution

In the above question we need to determine the angle between 2a⃗+b⃗2\vec a + \vec b and a⃗+2b⃗\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 2a⃗+b⃗2\vec a + \vec b and a⃗+2b⃗\vec a + 2\vec b such that the value are
a⃗=i^+2j^−3k^\vec a = \hat i + 2\hat j - 3\hat k
And
b⃗=3i^−j^+2k^\vec b = 3\hat i - \hat j + 2\hat k
Now substituting the above value we get
⇒\Rightarrow 2a⃗+b⃗=2(i^+2j^−3k^)+3i^−j^+2k^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 2a⃗+b⃗=2i^+4j^−6k^+3i^−j^+2k^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 2a⃗+b⃗=5i^+3j^−4k^2\vec a + \vec b = 5\hat i + 3\hat j - 4\hat k
Now, we will determine the value of
⇒\Rightarrow a⃗+2b⃗=i^+2j^−3k^+2(3i^−j^+2k^)\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 a⃗+2b⃗=i^+2j^−3k^+2(3i^−j^+2k^)\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 a⃗+2b⃗=i^+2j^−3k^+6i^−2j^+4k^\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 a⃗+2b⃗=7i^+k^\vec a + 2\vec b = 7\hat i + \hat k
Now, let us determine the magnitude of both , we get
⇒\Rightarrow ∣2a⃗+b⃗∣=(5)2+(3)2+(−4)2=25+9+16=50|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,onsubstitutingtherequiredvalueweget 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 Onsimplifyingtheabove,weget On simplifying the above, we get \Rightarrow 35 - 4 = 50\cos \theta Hence,weget Hence, we get \Rightarrow \dfrac{{31}}{{50}} = \cos \theta Now,weneedtodeterminetheangle 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(3150){\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 i^\hat i with j^,k^\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 3131 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.