What is the magnitude of vector, v = 1/√3 i + 1/√3 j + 1/√3 k?A3B2C1D0

To find the magnitude of the vector $\mathbf{v} = \frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k}$, we can use the formula for the magnitude of a vector in three-dimensional space:

$\|\mathbf{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2}$

where $v_x$, $v_y$, and $v_z$ are the components of the vector along the $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ directions, respectively.

Given: $v_x = \frac{1}{\sqrt{3}}$ $v_y = \frac{1}{\sqrt{3}}$ $v_z = \frac{1}{\sqrt{3}}$

Now, substitute these values into the magnitude formula:

$\|\mathbf{v}\| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2}$

Calculate each term inside the square root:

$\left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$

So, we have:

$\|\mathbf{v}\| = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}}$

Combine the terms inside the square root:

$\|\mathbf{v}\| = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = \sqrt{\frac{3}{3}} = \sqrt{1}$

Finally, we get:

$\|\mathbf{v}\| = 1$

Therefore, the magnitude of the vector $\mathbf{v}$ is 1.

The correct answer is: C) 1