Question

if two unit vectors are perpendicular to each other are added, how their resultant will be square root of 2
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Answer 1

The answer is the square root of 2, which equals approximately 1.414.

You can think of the two perpendicular vectors as the perpendicular sides of a right triangle. The vector formed by adding them together is the third side - the hypotenuse.

The pythagorean equation is, c^2 = a^2 + b^2, where “a” and “b” are the lengths of the perpedicular sides, and ”c” is the hypotenuse.

So since our perpendicular vectors are also unit vectors, we know that both of their lengths (magnitudes) are 1 unit. We can plug those 1s into the pythagorean equation, and solve for the length of the hypotenuse, which is “c” in the equation, and also the magnitude of our summed vector.

c^2 = 1^2 + 1^2

One times one is still one, so simplifying the 1^2s:

c^2 = 1 + 1

Adding the numbers together:

c^2 = 2

Taking the square root of both sides:

c = sqrt(2)

Answer 2

Answer:

It can be proved considering two perpendicular vectors as the perpendicular sides of a right triangle.

Explanation:

It is given that two unit vectors are perpendicular to each other.

We need to prove that if these are added, how their resultant will be square root of $2$.

We can consider two perpendicular vectors as the perpendicular sides of a right triangle.

The vector formed by adding them together is the third side - the hypotenuse.

We know that, pythagorean equation is

$c^2 = a^2 + b^2$

$a,b$ - lengths of perpendicular

$c$ is hypotenuse.

As perpendicular vectors are also unit vectors, we know that both of their lengths are $1$ unit.

Using these values, we get

$c^2 = 1^2 + 1^2$

$c^2 = 1+1=2$

Taking the square root of both sides:

$c = \sqrt{2}$

Hence proved.

#SPJ2