Question

the sum of two unit vector is also a unit vector.then magnitude of their different is..
1) 0. 2) root 2. 3) root 3. 4) root 7

Answer 1

Answer on Question 51622, Physics, Mechanics | Kinematics | Dynamics

Question:

The sum of two unit vector is also a unit vector. Then magnitude of their difference is

1) 0

2) $\sqrt{2}$

3) $\sqrt{3}$

4) $\sqrt{7}$

Solution:

Let $\mathbf{a}$ and $\mathbf{b}$ be a unit vectors, therefore their magnitudes will be $\| \mathbf{a} \| = \| \mathbf{b} \| = 1$. Because $\mathbf{a} + \mathbf{b}$ has unit length we get:

$$1 = \| \mathbf{a} + \mathbf{b} \|^2,$$ $$1 = \| \mathbf{a} \|^2 + \| \mathbf{b} \|^2 + 2 \langle \mathbf{a}, \mathbf{b} \rangle,$$ $$1 = 2 + 2 \langle \mathbf{a}, \mathbf{b} \rangle,$$ $$\langle \mathbf{a}, \mathbf{b} \rangle = -\frac{1}{2}.$$

Here, $\langle \mathbf{a}, \mathbf{b} \rangle$ is the dot product or a scalar product of two vectors $\mathbf{a}$ and $\mathbf{b}$, and we need it in order to obtain the magnitude of their difference:

$$\| \mathbf{a} - \mathbf{b} \|^2 = \| \mathbf{a} \|^2 + \| \mathbf{b} \|^2 - 2 \langle \mathbf{a}, \mathbf{b} \rangle,$$ $$\| \mathbf{a} - \mathbf{b} \|^2 = 2 - 2 \cdot \left(-\frac{1}{2}\right),$$ $$\| \mathbf{a} - \mathbf{b} \|^2 = 3,$$ $$\| \mathbf{a} - \mathbf{b} \| = \sqrt{3}.$$

Therefore, we get that the magnitude of their difference is $\sqrt{3}$.

Answer:

3) $\sqrt{3}$

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