Question

If we are given three vectors $\vec{a},\vec{b},\vec{c}$ such that $\left| \overrightarrow{a} \right|=3,\left| \overrightarrow{b} \right|=4,\left| \overrightarrow{c} \right|=5$ and $\vec{a},\vec{b},\vec{c}$ are perpendicular to $\vec{b}+\vec{c},\vec{c}+\vec{a},\vec{a}+\vec{b}$ respectively, then $\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|=?$
1. $5\sqrt{2}$
2. $6\sqrt{2}$
3. $3\sqrt{2}$
4. $4\sqrt{2}$

Answer 1

Now to solve this question we must use the logic of vectors that we know that when two vectors are perpendicular to each other then their dot product will be zero. Here we also use the formula for vectors that
$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{{|}^{2}}={{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})$ .

Complete step-by-step solution:
To solve this question;
We know that here in this case $\overrightarrow{a}\bot (\vec{b}+\vec{c})$
Now since we know that when two vectors are perpendicular it means that their dot product will always be equal to zero. This means that
$\vec{a}.(\vec{b}+\vec{c})=0$
Now when we open the bracket and take the dot product we get
$\vec{a}.\vec{b}+\vec{a}.\vec{c}=0$ ---- $1$
Now take this as equation first
Similarly we can use this same logic for other vectors and we get;
$\vec{b}\bot (\vec{c}+\vec{a})$
Now since we know that when two vectors are perpendicular it means that their dot product will always be equal to zero. This means that
$\vec{b}.(\vec{c}+\vec{a})=0$
Now when we open the bracket and take the dot product we get
$\vec{b}.\vec{c}+\vec{b}.\vec{a}=0$ ---- $2$
Lastly
$\vec{c}\bot (\vec{a}+\vec{b})$
Now since we know that when two vectors are perpendicular it means that their dot product will always be equal to zero. This means that
$\vec{c}.(\vec{a}+\vec{b})=0$
Now when we open the bracket and take the dot product we get
$\vec{c}.\vec{a}+\vec{c}.\vec{b}=0$ ---- $3$
Now adding all three equations we get that
$2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=0$
Now here in this question it is asked of us that we need to find $\left| \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \right|=?$. Now the formula of this expression is
$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{{|}^{2}}={{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})$
Now here we know that $2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=0$ as derived in the solution above and we also know the magnitude of vectors.
Substituting we get
$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{{|}^{2}}={{3}^{2}}+{{4}^{2}}+{{5}^{2}}+0$
Taking the squares and adding
$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{{|}^{2}}=50$
Taking square root of both sides
$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=5\sqrt{2}$
Hence the answer for this question is option (1).

Note: Now here in this question the main concept being used is the dot product of two vectors. Students must note that the dot product of two perpendicular vectors will always be equal to zero. If students won’t know these concepts they wouldn’t be able to solve the above question and questions like this. Also a common mistake is that usually the formula is mistaken to be $|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|={{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})$ instead of $|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{{|}^{2}}={{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})$ , which gives us the wrong answer.