Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular vectors such that $\left| \overrightarrow{a}+\overrightarrow{b} \right|=13$ and $\left| \overrightarrow{a} \right|=5$, find the value of $\left| \overrightarrow{b} \right|$.

Answer 1

Hint: To solve this question, what we will do is, we first find out the value of $\overrightarrow{a}\cdot \overrightarrow{b}$, when $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular vectors, then we will square the $\left| \overrightarrow{a}+\overrightarrow{b} \right|=13$ on both the sides and then we will substitute the values of $\overrightarrow{a}\cdot \overrightarrow{b}$ and $\left| \overrightarrow{a} \right|=5$ to obtain the value of $\left| \overrightarrow{b} \right|$.

Complete step by step answer:
Now if it is given that two vectors, say $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular vectors then always remember that $\overrightarrow{a}\cdot \overrightarrow{b}=0$.
Now, in question it is given that, $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular vectors such that $\left| \overrightarrow{a}+\overrightarrow{b} \right|=13$ and $\left| \overrightarrow{a} \right|=5$
Now, let us solve, $\left| \overrightarrow{a}+\overrightarrow{b} \right|=13$ first.
$\left| \overrightarrow{a}+\overrightarrow{b} \right|=13$
Squaring both sides, we get
${{\left| \overrightarrow{a}+\overrightarrow{b} \right|}^{2}}={{13}^{2}}$
$\left( \overrightarrow{a}+\overrightarrow{b} \right)\cdot \left( \overrightarrow{a}+\overrightarrow{b} \right)={{13}^{2}}$
On, solving brackets, we get
${{\left| \overrightarrow{a} \right|}^{2}}+2\cdot \overrightarrow{a\cdot }\overrightarrow{b}+{{\left| \overrightarrow{b} \right|}^{2}}={{13}^{2}}$.
Now, as in question it is given that $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular vectors and we discussed above that, $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular vectors then, $\overrightarrow{a}\cdot \overrightarrow{b}=0$.
Also, here $\overrightarrow{a}\cdot \overrightarrow{b}=ab\cos \theta $ which is called as dot product of vector $\overrightarrow{a}$ and $\overrightarrow{b}$.
So, putting $\overrightarrow{a}\cdot \overrightarrow{b}=0$in ${{\left| \overrightarrow{a} \right|}^{2}}+2\cdot \overrightarrow{a\cdot }\overrightarrow{b}+{{\left| \overrightarrow{b} \right|}^{2}}={{13}^{2}}$, we get
${{\left| \overrightarrow{a} \right|}^{2}}+2\cdot (0)+{{\left| \overrightarrow{b} \right|}^{2}}={{13}^{2}}$,
On solving, we get
${{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}=169$ .
Also, in question it is given that $\left| \overrightarrow{a} \right|=5$,
So, squaring both side, we get
 ${{\left| \overrightarrow{a} \right|}^{2}}={{5}^{2}}$ .
${{\left| \overrightarrow{a} \right|}^{2}}=25$ .
Putting, ${{\left| \overrightarrow{a} \right|}^{2}}=25$in ${{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}=169$, we get
$25+{{\left| \overrightarrow{b} \right|}^{2}}=169$ .
Taking, 25 from left hand side to right hand side, we get
${{\left| \overrightarrow{b} \right|}^{2}}=169-25$ .
On simplifying, we get
${{\left| \overrightarrow{b} \right|}^{2}}=144$ .
Taking square root on both side, we get
$\left| \overrightarrow{b} \right|=\sqrt{144}$ .
$\left| \overrightarrow{b} \right|=12$ .

Hence, value of $\left| \overrightarrow{b} \right|$is equals to $\left| \overrightarrow{b} \right|=\sqrt{144}$.

Note: vectors are an important portion of mathematics, so one must know basics, theory and all concepts of vectors. While solving questions related to vectors, calculation error must be avoided and always represent a vector by an arrow on its head. Vector is something which has magnitude and direction both.