Question #b2c5e

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Answer 1 · 2017-11-10T07:55:13

See the proof below

Let $vecu= < a, b,c >$ in the basis $beta=(hati, hatj, hatk)$

and $vecv= < p,q,r >$

It is given that $||vecu|| = ||vecv||$

Then

$vecu+vecv=< a, b,c > + < p,q,r >$

$=< a+p,b+q,c+r >$

$vecu-vecv=< a, b,c > - < p,q,r >$

$=< a-p, b-q, c-r >$

The dot product is

$(vecu+vecv).(vecu-vecv)$

$=< a+p,b+q,c+r > . < a-p, b-q, c-r >$

$= (a+p)(a-p) +(b+q)(b-q) + (c+r)(c-r)$

$=a^2-p^2+b^2-q^2+c^2-r^2$

$=(a^2+b^2+c^2) - (p^2+q^2+r^2)$

$=||vecu||^2- ||vecv||^2=0$

Therefore,

$(vecu+vecv)$ and $(vecu-vecv)$ are orthogonal since their dots products $=0$

Answer 2 · 2017-11-10T11:30:01

Consider two vectors $vec u$ and $vec v$ .

Let $vec w = vec u + vec v$ and $vec x = vec u - vec v$

Now, $vecw*vecw = u^2 + v^2 + 2uvcos theta$

Similarly,
$vecx*vecx = u^2 + v^2 - 2uvCos theta$ where $theta$ is the angle between vectors $vec u$ and $vec v$ .

It is given that $|w| = |x|$
$implies w^2 = x^2$
$implies vecw*vecw = vecx*vecx$
$implies u^2 + v^2 + 2uvcos theta = u^2 + v^2 -2uvcos theta$

Thus, we may obtain $4uvcos theta = 0$ but for $u$ and $v$ non zero, therefore $cos theta = 0$ and hence $vec u$ and $vec v$ are orthogonal.