Question

If $a,bandc$are three vectors such that $\left|a\right|=3,\left|b\right|=4\&\left|c\right|=5$ and$a,bandc$ are perpendicular to $b+c,c+a\&a+b$respectively, then $\left|a+b+c\right|=?$

• A. $5\sqrt{2}$
• B. $50$
• C. $10\sqrt{2}$
• D. $10$

Answer 1

The correct option is A

$5\sqrt{2}$

Explanation for the correct option:

Step 1: Find the value of $(a.b+b.c+a.c)$.

We have given$\left|a\right|=3,\left|b\right|=4\&\left|c\right|=5$ and$a,bandc$ are perpendicular to $b+c,c+a\&a+b$ respectively.

We need to find the$\left|a+b+c\right|=?$

Since, $\begin{array}{rcl}& & \mathrm{a}\perp \mathrm{b}+\mathrm{c}\\ \therefore a.(b+c)& =& 0\\ a.b+a.c\mathbf{}\mathbf{}\mathbf{}& =& 0.......\left(1\right)\end{array}$

similarly, $\begin{array}{rcl}& & \mathrm{b}\perp \mathrm{c}+\mathrm{a}\\ \therefore b.(c+a)\mathbf{}& =& 0\\ b.c+b.a\mathbf{}\mathbf{}& =& 0.......\left(2\right)\end{array}$

and $\begin{array}{rcl}& & \mathrm{c}\perp \mathrm{a}+\mathrm{b}\\ \therefore c.(a+b)\mathbf{}& =& 0\\ c.a+c.b\mathbf{}\mathbf{}\mathbf{}& =& 0.......\left(3\right)\end{array}$

adding$\left(1\right),\left(2\right)\&\left(3\right)$,

$$\begin{gathered} 2(a.b+b.c+a.c)=0 \\ (a.b+b.c+a.c)=0 \end{gathered}$$

Step 2: Take the dot product of$\begin{array}{l}(a+b+c)\end{array}$ with itself.

${\left|a+b+c\right|}^2=$$\begin{array}{rcl}(a+b+c).(a+b+c)& =& {\left|a\right|}^2+{\left|b\right|}^2+{\left|c\right|}^2+2(a.b+b.c+a.c)\end{array}$

From above$\begin{array}{rcl}(a.b+b.c+a.c)& =& 0\end{array}$

$$\begin{gathered} \Rightarrow {|\mathrm{a}+\mathrm{b}+\mathrm{c}|}^2=3^2+4^2+5^2 \\ \Rightarrow |\mathrm{a}+\mathrm{b}+\mathrm{c}{|}^2=50 \\ \Rightarrow |\mathrm{a}+\mathrm{b}+\mathrm{c}|=\sqrt{50} \\ \Rightarrow |\mathrm{a}+\mathrm{b}+\mathrm{c}|=5\sqrt{2} \end{gathered}$$

Hence, the option$\left(A\right)$ is the correct option.