Question

Please solve this asap​

Answer 1

Let $\vec u$ and $\vec v$ be the vectors, and let $x=\|\vec u\|=\|\vec v\|$ be their common magnitude.

The resultant $\vec u + \vec v$ is $\sqrt 2$ times larger in magnitude than either vector alone, so $\|\vec u+\vec v\| = \sqrt2\,x$.

Recall the dot product identity

$\vec a \cdot \vec b = \|\vec a\| \|\vec b\| \cos(\theta)$

where $\theta$ is the angle between the vectors $\vec a$ and $\vec b$. In the special case of $\vec a=\vec b$, we get

$\vec a \cdot \vec a = \|\vec a\|^2 \cos(0^\circ) \implies \|\vec a\| = \sqrt{\vec a\cdot\vec a}$

Now, to get the angle between $\vec u$ and $\vec v$, we have

$\vec u \cdot \vec v = \|\vec u\| \|\vec v\| \cos(\theta) \implies \cos(\theta) = \dfrac{\vec u \cdot \vec v}{x^2}$

To compute the dot product, we take the dot product of the resultant with itself.

$(\vec u+\vec v) \cdot (\vec u + \vec v) = \|\vec u + \vec v\|^2$

Solve for $\vec u\cdot\vec v$.

$(\vec u\cdot\vec u) + 2(\vec u\cdot\vec v) + (\vec v\cdot\vec v) = \|\vec u + \vec v\|^2$

$\|\vec u\|^2 + 2(\vec u\cdot \vec v) + \|\vec v\|^2 = \|\vec u+\vec v\|^2$

$x^2 + 2(\vec u \cdot \vec v) + x^2 = (\sqrt2\,x)^2$

$2(\vec u\cdot\vec v) + 2x^2 = 2x^2$

$2(\vec u\cdot\vec v) = 0$

$\vec u\cdot\vec v = 0$

Since their dot product is zero, $\vec u$ and $\vec v$ are perpendicular, so $\theta=90^\circ$.