Question

If $\hat{a},\hat{b}$ and $\hat{c}$ are mutually perpendicular unit vectors, then find the value of $|2\hat{a}+\hat{b}+\hat{c}|$

Answer 1

Now we are given that If $\hat{a},\hat{b}$ and $\hat{c}$ are mutually perpendicular unit vectors $\hat{a},\hat{b}$ and $\hat{c}$ are mutually perpendicular vectors and dot product of perpendicular vectors is equal to zero.
And the modulus of each vector is equal to 1. Then we will consider $|2\hat{a}+\hat{b}+\hat{c}{{|}^{2}}$. We know that $\hat{a}.\hat{a}=|\hat{a}{{|}^{2}}$ . Hence using this property we get the value of $|2\hat{a}+\hat{b}+\hat{c}{{|}^{2}}$ in this we will substitute 0 and 1 as per the conditions we get from the fact that the vectors are perpendicular vectors and unit vectors. Hence once we find the value of $|2\hat{a}+\hat{b}+\hat{c}{{|}^{2}}$ we will take square root on both sides to get the value of $|2\hat{a}+\hat{b}+\hat{c}|$

Complete step-by-step answer:
Now we know that since $\hat{a},\hat{b}$ and $\hat{c}$ are mutually perpendicular vectors and dot product of perpendicular vectors is equal to zero.
$\hat{a}.\hat{b}=\hat{b}.\hat{c}=\hat{c}.\hat{a}=0..........(1)$
And also since $\hat{a},\hat{b}$ and $\hat{c}$ are unit vector we have
$|\hat{a}|=|\hat{b}|=|\hat{c}|=1...........(2)$
Now let us consider $|2\hat{a}+\hat{b}+\hat{c}{{|}^{2}}$
Now we know that $\hat{a}.\hat{a}=|\hat{a}{{|}^{2}}$ hence we get
$|2\hat{a}+\hat{b}+\hat{c}{{|}^{2}}=\left( 2\hat{a}+\hat{b}+\hat{c} \right).\left( 2\hat{a}+\hat{b}+\hat{c} \right)$
Now we know that the dot product is distributive over addition which means $\hat{a}.(\hat{b}+\hat{c})=\hat{a}.\hat{b}+\hat{a}.\hat{c}$ . Hence using this property we get
$|2\hat{a}+\hat{b}+\hat{c}{{|}^{2}}=4(\hat{a}.\hat{a})+2\left( \hat{a}.\hat{b}. \right)+2\left( \hat{a}.\hat{c} \right)+2\left( \hat{b}.\hat{a} \right)+(\hat{b}.\hat{b})+\left( \hat{b}.\hat{c} \right)+2\left( \hat{c}.\hat{a} \right)+2\left( \hat{c}.\hat{b} \right)+\left( \hat{c}.\hat{c} \right)$
Now we know that $\hat{a}.\hat{a}=|\hat{a}{{|}^{2}}$
Hence using this in the above equation we get
$|2\hat{a}+\hat{b}+\hat{c}{{|}^{2}}=4|\hat{a}{{|}^{2}}+2\left( \hat{a}.\hat{b}. \right)+2\left( \hat{a}.\hat{c} \right)+2\left( \hat{b}.\hat{a} \right)+|\hat{b}{{|}^{2}}+\left( \hat{b}.\hat{c} \right)+2\left( \hat{c}.\hat{a} \right)+2\left( \hat{c}.\hat{b} \right)+|\hat{c}{{|}^{2}}$
Now substituting the values from equation (2) and equation (1) we get
$|2\hat{a}+\hat{b}+\hat{c}{{|}^{2}}=4(1)+2\left( 0. \right)+2\left( 0 \right)+2(0)+(1)+\left( 0 \right)+2\left( 0 \right)+2\left( 0 \right)+\left( 1 \right)$
Adding the terms on RHS we get
$|2\hat{a}+\hat{b}+\hat{c}{{|}^{2}}=6$
Now let us take square root on both sides so that we get the value of $|2\hat{a}+\hat{b}+\hat{c}|$ . Hence we get.
$|2\hat{a}+\hat{b}+\hat{c}|=+\sqrt{6}$ or $|2\hat{a}+\hat{b}+\hat{c}|=-\sqrt{6}$
Now we know that modulus is always positive hence we will eliminate the case of negative answer
Hence we get the answer as $|2\hat{a}+\hat{b}+\hat{c}|=\sqrt{6}$

Note: Note dot product of two perpendicular vectors is 0 and not 1. Also we have $\hat{a}.\hat{a}=|\hat{a}{{|}^{2}}$ and not $\hat{a}.\hat{a}=|\hat{a}|$.