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\| = √(\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) = (\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`.