Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ is equally inclined to $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$.

Answer 1

Given, $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$
($\because$ a, b, c are mutually perpendicular)
It is also given that |a| = |b| = |c| ($\because$ All the vectors have same magnitude)
Let vector $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ be inclined to $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ at angles $\alpha ,\beta$ and $\gamma$ respectively.
Then, we have
$cos\alpha =\frac{(a+b+c).a}{|a+b+c||a|}=\frac{a.a+b.a+c.a}{|a+b+c||a|}=\frac{a.a+0+0}{|a+b+c||a|}=\frac{|a{|}^2}{|a+b+c||a|}$ $(\because b.c=c.a=0)$
$=\frac{|a|}{|a+b+c|}$

$cos\beta =\frac{(a+b+c).b}{|a+b+c||a|}=\frac{a.b+b.b+c.b}{|a+b+c||b|}$
$=\frac{b.b+0+0}{|a+b+c||b|}=\frac{|b{|}^2}{|a+b+c||b|}(\because a.b=c.b=0)=\frac{|b|}{|a+b+c|}$

$cos\gamma =\frac{(a+b+c).c}{|a+b+c||c|}=\frac{a.c+b.c+c.c}{|a+b+c||c|}=\frac{c.c+0+0}{|a+b+c||c|}=\frac{|c{|}^2}{|a+b+c||c|}$

$(\because a.c=b.c=0)$
$=\frac{|c|}{|a+b+c|}$.
Now, as |a| = |b| = |c|, therefore, $cos\alpha =cos\beta =cos\gamma$
$\therefore \alpha =\beta =\gamma$
Hence, the vector (a + b + c) is equally inclined to a, b and c.