
If $\vec a = \hat i + 2\hat j - 3\hat k$ and $\vec b = 3\hat i - \hat j + 2\hat k$, then the angle between the vectors $\vec a + \vec b$ and $\vec a - \vec b$?
(A) \[{60^ \circ }\]
(B) \[{90^ \circ }\]
(C) \[{45^ \circ }\]
(D) \[{55^ \circ }\]
Answer
232.8k+ views
Hint: We are provided with the vectors $\vec a = \hat i + 2\hat j - 3\hat k$ and $\vec b = 3\hat i - \hat j + 2\hat k$, and we have to find the angle between the vectors formed by addition and subtraction of the two vectors. So, we first find the sum and difference of the vectors and then find their dot product to get the angle between the resultant vectors.
Formula used: $\vec x \cdot \vec y = \left| {\vec x} \right|\left| {\vec y} \right|\cos \theta $
Complete step by step solution:
We have, $\vec a = \hat i + 2\hat j - 3\hat k$ and $\vec b = 3\hat i - \hat j + 2\hat k$.
Now, \[\vec a + \vec b = \left( {\hat i + 2\hat j - 3\hat k} \right) + \left( {3\hat i - \hat j + 2\hat k} \right)\]
\[ \Rightarrow \vec a + \vec b = 4\hat i + \hat j - \hat k\]
Also, \[\vec a - \vec b = \left( {\hat i + 2\hat j - 3\hat k} \right) - \left( {3\hat i - \hat j + 2\hat k} \right)\]
\[ \Rightarrow \vec a - \vec b = - 2\hat i + 3\hat j - 5\hat k\]
Now, we have to find the angle between the vectors $\vec a + \vec b$ and $\vec a - \vec b$.
So, $\left( {\vec a + \vec b} \right) \cdot \left( {\vec a - \vec b} \right) = \left( {4\hat i + \hat j - \hat k} \right) \cdot \left( { - 2\hat i + 3\hat j - 5\hat k} \right)$
\[ \Rightarrow \left( {\vec a + \vec b} \right) \cdot \left( {\vec a - \vec b} \right) = 4 \times \left( { - 2} \right) + 1 \times 3 + \left( { - 1} \right) \times \left( { - 5} \right)\]
$ \Rightarrow \left( {\vec a + \vec b} \right) \cdot \left( {\vec a - \vec b} \right) = - 8 + 3 + 5$
$ \Rightarrow \left( {\vec a + \vec b} \right) \cdot \left( {\vec a - \vec b} \right) = 0$
We know the formula of dot product of two vectors as, $\vec x \cdot \vec y = \left| {\vec x} \right|\left| {\vec y} \right|\cos \theta $, where the angle between the two vectors is $\theta $.
Since the dot product is zero, we get,
$ \Rightarrow \left| {\vec a + \vec b} \right|\left| {\vec a - \vec b} \right|\cos \theta = 0$
Now, cosine of the angle between the vectors must be zero, since the magnitudes of the vectors is non-zero. So, we get,
$ \Rightarrow \cos \theta = 0$
Now, cosine of a right angle is zero. So, we get,
$ \Rightarrow \theta = {90^ \circ }$
Hence, option (B) is correct.
Note: The given question revolves around the concepts of the addition of two vectors and the angle between two vectors. The angle between the two vectors can be calculated using the dot product of two vectors as the formula of the scalar product has a cosine of the angle between the two vectors in it. The method of calculating the dot product of two vectors in rectangular coordinates should be remembered as it is of vital use in solving such questions.
Formula used: $\vec x \cdot \vec y = \left| {\vec x} \right|\left| {\vec y} \right|\cos \theta $
Complete step by step solution:
We have, $\vec a = \hat i + 2\hat j - 3\hat k$ and $\vec b = 3\hat i - \hat j + 2\hat k$.
Now, \[\vec a + \vec b = \left( {\hat i + 2\hat j - 3\hat k} \right) + \left( {3\hat i - \hat j + 2\hat k} \right)\]
\[ \Rightarrow \vec a + \vec b = 4\hat i + \hat j - \hat k\]
Also, \[\vec a - \vec b = \left( {\hat i + 2\hat j - 3\hat k} \right) - \left( {3\hat i - \hat j + 2\hat k} \right)\]
\[ \Rightarrow \vec a - \vec b = - 2\hat i + 3\hat j - 5\hat k\]
Now, we have to find the angle between the vectors $\vec a + \vec b$ and $\vec a - \vec b$.
So, $\left( {\vec a + \vec b} \right) \cdot \left( {\vec a - \vec b} \right) = \left( {4\hat i + \hat j - \hat k} \right) \cdot \left( { - 2\hat i + 3\hat j - 5\hat k} \right)$
\[ \Rightarrow \left( {\vec a + \vec b} \right) \cdot \left( {\vec a - \vec b} \right) = 4 \times \left( { - 2} \right) + 1 \times 3 + \left( { - 1} \right) \times \left( { - 5} \right)\]
$ \Rightarrow \left( {\vec a + \vec b} \right) \cdot \left( {\vec a - \vec b} \right) = - 8 + 3 + 5$
$ \Rightarrow \left( {\vec a + \vec b} \right) \cdot \left( {\vec a - \vec b} \right) = 0$
We know the formula of dot product of two vectors as, $\vec x \cdot \vec y = \left| {\vec x} \right|\left| {\vec y} \right|\cos \theta $, where the angle between the two vectors is $\theta $.
Since the dot product is zero, we get,
$ \Rightarrow \left| {\vec a + \vec b} \right|\left| {\vec a - \vec b} \right|\cos \theta = 0$
Now, cosine of the angle between the vectors must be zero, since the magnitudes of the vectors is non-zero. So, we get,
$ \Rightarrow \cos \theta = 0$
Now, cosine of a right angle is zero. So, we get,
$ \Rightarrow \theta = {90^ \circ }$
Hence, option (B) is correct.
Note: The given question revolves around the concepts of the addition of two vectors and the angle between two vectors. The angle between the two vectors can be calculated using the dot product of two vectors as the formula of the scalar product has a cosine of the angle between the two vectors in it. The method of calculating the dot product of two vectors in rectangular coordinates should be remembered as it is of vital use in solving such questions.
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