Answer
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Hint : To solve this question, we will use the concept of properties of scalar products of vectors. According to that, if \[\hat i,\hat j,\hat k\] are three mutually perpendicular unit vectors along the coordinates axes, then \[\hat i.\hat j = \hat j.\hat i = 0;\] \[\hat j.\hat k = \hat k.\hat j = 0;\] \[\hat k.\hat i = \hat i.\hat k = 0\].
Complete step by step solution:
A vector whose magnitude in unity, is called a unit vector. The unit vector in the direction of a vector $\overrightarrow a $ is denoted by $\hat a$. Thus, $\left| {\hat a} \right| = 1$.
Given that, $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are three mutually perpendicular unit vectors.
We have to prove that $\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 3 $.
Proof :
As we know that if \[\hat i,\hat j,\hat k\] are three mutually perpendicular unit vectors along the coordinates axes, then \[\hat i.\hat j = \hat j.\hat k = \hat k.\hat i = 0\].
Since, $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are three mutually perpendicular unit vectors, therefore
$\overrightarrow a .\overrightarrow b = \overrightarrow b .\overrightarrow c = \overrightarrow c .\overrightarrow a = 0$. ……….. (i)
We know that, ${\left| {\overrightarrow x } \right|^2} = \overrightarrow x .\overrightarrow x $.
Therefore,
${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right).\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)$.
$
\Rightarrow {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = \overrightarrow a .\overrightarrow a + \overrightarrow b .\overrightarrow b + \overrightarrow c .\overrightarrow c + 2\overrightarrow a .\overrightarrow b + 2\overrightarrow b .\overrightarrow c + 2\overrightarrow c .\overrightarrow a \\
\Rightarrow {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = \overrightarrow a .\overrightarrow a + \overrightarrow b .\overrightarrow b + \overrightarrow c .\overrightarrow c + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right) \\
$
Putting the value of $\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a $ from equation (i), we will get
$ \Rightarrow {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2}$ ……….. (ii)
According to the question, $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are unit vectors and we know that the magnitude of a unit vector is always equal to 1.
Therefore,
${\left| {\overrightarrow a } \right|^2} = {\left| {\overrightarrow b } \right|^2} = {\left| {\overrightarrow c } \right|^2} = 1$ [ $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are unit vectors ]
Putting this value in equation (ii), we will get
$ \Rightarrow {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = 1 + 1 + 1$
$ \Rightarrow {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = 3$
Taking square root both sides, we will get
$ \Rightarrow \left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 3 $.
Hence proved, If $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are three mutually perpendicular unit vectors, then we can say that $\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 3 $.
Note : whenever we ask such types of questions, we have to remember the properties of the scalar product of vectors. First we have to find out what is given in the question then what we have to prove. After that we will use the properties on the given part. We will use the concept of unit vectors and mutually perpendicular vectors. Using this, we can easily prove the question and we will get the answer.
Complete step by step solution:
A vector whose magnitude in unity, is called a unit vector. The unit vector in the direction of a vector $\overrightarrow a $ is denoted by $\hat a$. Thus, $\left| {\hat a} \right| = 1$.
Given that, $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are three mutually perpendicular unit vectors.
We have to prove that $\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 3 $.
Proof :
As we know that if \[\hat i,\hat j,\hat k\] are three mutually perpendicular unit vectors along the coordinates axes, then \[\hat i.\hat j = \hat j.\hat k = \hat k.\hat i = 0\].
Since, $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are three mutually perpendicular unit vectors, therefore
$\overrightarrow a .\overrightarrow b = \overrightarrow b .\overrightarrow c = \overrightarrow c .\overrightarrow a = 0$. ……….. (i)
We know that, ${\left| {\overrightarrow x } \right|^2} = \overrightarrow x .\overrightarrow x $.
Therefore,
${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right).\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)$.
$
\Rightarrow {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = \overrightarrow a .\overrightarrow a + \overrightarrow b .\overrightarrow b + \overrightarrow c .\overrightarrow c + 2\overrightarrow a .\overrightarrow b + 2\overrightarrow b .\overrightarrow c + 2\overrightarrow c .\overrightarrow a \\
\Rightarrow {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = \overrightarrow a .\overrightarrow a + \overrightarrow b .\overrightarrow b + \overrightarrow c .\overrightarrow c + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right) \\
$
Putting the value of $\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a $ from equation (i), we will get
$ \Rightarrow {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2}$ ……….. (ii)
According to the question, $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are unit vectors and we know that the magnitude of a unit vector is always equal to 1.
Therefore,
${\left| {\overrightarrow a } \right|^2} = {\left| {\overrightarrow b } \right|^2} = {\left| {\overrightarrow c } \right|^2} = 1$ [ $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are unit vectors ]
Putting this value in equation (ii), we will get
$ \Rightarrow {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = 1 + 1 + 1$
$ \Rightarrow {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = 3$
Taking square root both sides, we will get
$ \Rightarrow \left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 3 $.
Hence proved, If $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are three mutually perpendicular unit vectors, then we can say that $\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 3 $.
Note : whenever we ask such types of questions, we have to remember the properties of the scalar product of vectors. First we have to find out what is given in the question then what we have to prove. After that we will use the properties on the given part. We will use the concept of unit vectors and mutually perpendicular vectors. Using this, we can easily prove the question and we will get the answer.
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