Question

Write the vector m/s² as the product $\left|\stackrel{\rightharpoonup }{a}\right|-\stackrel{\rightharpoonup }{a}$.

Short answer

The required product is .

Step-by-step solution

Step 1: Definition of the magnitude of a vector 

The magnitude of a vectorhaving coordinates is written as:.

$\left|\stackrel{\rightharpoonup }{a}\right|=\sqrt{x^2+y^2+z^2}$

The unit vector in the direction of a vector is the vector divided by its magnitude, which is given by:

.$\stackrel{\rightharpoonup }{a}=\frac{\stackrel{\rightharpoonup }{a}}{\stackrel{\rightharpoonup }{a}}$

Calculation of magnitude of the vector

Substitute $x=400,y=200andz=100$into the formula of magnitude to get the magnitude of the vector .

$$\begin{gathered} \left|\stackrel{\rightharpoonup }{a}\right|=\sqrt{{400}^2+{200}^2+{\left(-100\right)}^2} \\ =100\sqrt{21} \end{gathered}$$

Calculating unit vector

Divide the given vector by its magnitude to get its unit vector.

The unit vector in the direction of a vector is the vector divided by its magnitude:

Calculating the Product of magnitude and the unit vector

Multiply the obtained magnitude by the obtained unit vector to get the required result.

Thus, the answer .