Question

Find $\left| \overset{\to }{\mathop{x}}\, \right|$ if for a unit vector $\overset{\to }{\mathop{a}}\,$, $\left( \overset{\to }{\mathop{x}}\,-\overset{\to }{\mathop{a}}\, \right).\left( \overset{\to }{\mathop{x}}\,+\overset{\to }{\mathop{a}}\, \right)=15$

Answer 1

Hint: Since we have the given expression $\left( \overset{\to }{\mathop{x}}\,-\overset{\to }{\mathop{a}}\, \right).\left( \overset{\to }{\mathop{x}}\,+\overset{\to }{\mathop{a}}\, \right)=15$, firstly we need to simplify the equation by solving the brackets and get a simple equation in terms of $\left| \overset{\to }{\mathop{x}}\, \right|$ and $\left| \overset{\to }{\mathop{a}}\, \right|$. Then, substitute the value of $\left| \overset{\to }{\mathop{a}}\, \right|$ in the equation and then get the value of $\left| \overset{\to }{\mathop{x}}\, \right|$.

Complete step by step answer:
We are given the following expression:
$\left( \overset{\to }{\mathop{x}}\,-\overset{\to }{\mathop{a}}\, \right).\left( \overset{\to }{\mathop{x}}\,+\overset{\to }{\mathop{a}}\, \right)=15......(1)$
Now, simplify equation (1) by opening the brackets and multiplying both the vectors, we get:
$\begin{align}
  & \Rightarrow \left( \overset{\to }{\mathop{x}}\,.\overset{\to }{\mathop{x}}\,+\overset{\to }{\mathop{x}}\,.\overset{\to }{\mathop{a}}\,-\overset{\to }{\mathop{x}}\,.\overset{\to }{\mathop{a}}\,-\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{a}}\, \right)=15 \\
 & \Rightarrow \left( \overset{\to }{\mathop{x}}\,.\overset{\to }{\mathop{x}}\,-\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{a}}\, \right)=15......(2) \\
\end{align}$
Since, we are squaring both the vectors, so square of a vector is equal to square of its magnitude, i.e. ${{\left( \overset{\to }{\mathop{x}}\, \right)}^{2}}={{\left| \overset{\to }{\mathop{x}}\, \right|}^{2}}$
So, we can write equation (2) as:
$\Rightarrow \left( {{\left| \overset{\to }{\mathop{x}}\, \right|}^{2}}-{{\left| \overset{\to }{\mathop{a}}\, \right|}^{2}} \right)=15......(3)$
Since it is mentioned in the question that $\overset{\to }{\mathop{a}}\,$ is a unit vector, therefore, we have $\left| \overset{\to }{\mathop{a}}\, \right|=1$
Now, by putting the value of $\left| \overset{\to }{\mathop{a}}\, \right|$in equation (2), we can write:
$\begin{align}
  & \Rightarrow {{\left| \overset{\to }{\mathop{x}}\, \right|}^{2}}-{{\left( 1 \right)}^{2}}=15 \\
 & \Rightarrow {{\left| \overset{\to }{\mathop{x}}\, \right|}^{2}}-1=15 \\
 & \Rightarrow {{\left| \overset{\to }{\mathop{x}}\, \right|}^{2}}=16 \\
 & \Rightarrow \left| \overset{\to }{\mathop{x}}\, \right|=4 \\
\end{align}$

Hence, the value of $\left| \overset{\to }{\mathop{x}}\, \right|$ is 4.

Note: As we have the expression $\left( \overset{\to }{\mathop{x}}\,-\overset{\to }{\mathop{a}}\, \right).\left( \overset{\to }{\mathop{x}}\,+\overset{\to }{\mathop{a}}\, \right)=15$ in the form of $\left( x-a \right)\left( x-b \right)$, we can apply the identity: ${{x}^{2}}-{{a}^{2}}=\left( x-a \right)\left( x-b \right)$.But instead of ${{\left( \overset{\to }{\mathop{x}}\, \right)}^{2}}$we need to write square of magnitude of the vector, i.e. ${{\left| \overset{\to }{\mathop{x}}\, \right|}^{2}}$. So, do not directly use the identity on the vectors, use it for the magnitude of vectors.
Also, when we get ${{\left| \overset{\to }{\mathop{x}}\, \right|}^{2}}=16$, we get two values of $\left| \overset{\to }{\mathop{x}}\, \right|$ i.e. +4 and -4. But magnitude cannot be negative. So, we have to neglect the negative value of $\left| \overset{\to }{\mathop{x}}\, \right|$, i.e. -4. Hence, we have $\left| \overset{\to }{\mathop{x}}\, \right|$ equal to 4.