Question

Figure below illustrates typical proportions of male (m) and female (f) anatomies. The displacements ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{1}\mathbf{m}}$ and ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{1}\mathbf{f}}$ from the soles of the feet to the navel have magnitude of 104 cm and 84.0 cm , respectively . The displacements ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{2}\mathbf{m}}$ and ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{2}\mathbf{f}}$ from the navel to outstretched fingertips have magnitude of 100 cm and 86.0 cm , respectively . Find the vector sum of these displacements ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{3}}\mathbf{=}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{1}}\mathbf{+}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{2}}$ for both people .

Short answer

For male magnitude = 170.1 cm and direction 57.2^o above +x axis

For female magnitude = 145.7 cm and direction 58.8^o above +x axis

Step-by-step solution

Definition of vector

A vector is a quantity that has both a direction and a magnitude, and is used to determine the relative location of two points in space.

Step 2:Formula

f vector A = a₁i + a₂j + a₃k and vector B = b₁i + b₂j + b₃k then the displacement of AB = (b₁ - a₁)i + (b₂ - a₂)j + (b₃ - a₃)k

If vector A = x₁i + y₁j + z₁k and vector B = x₂i + y₂j + z₂k then the displacement of

The Pythagorean theorem and the definition of tangent will use for calculation$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Calculation For Male

First, We sum the components of the two vector for the male

$$\begin{gathered} {\stackrel{\rightharpoonup }{d}}_{3mx}={\stackrel{\rightharpoonup }{d}}_{1mx}+{\stackrel{\rightharpoonup }{d}}_{2mx}=0+\left(100\times \cos {\left(23.0\right)}^o\right)=92.1cm \\ {\stackrel{\rightharpoonup }{d}}_{3my}={\stackrel{\rightharpoonup }{d}}_{1my}+{\stackrel{\rightharpoonup }{d}}_{2my}=104+\left(100\times \sin {\left(23.0\right)}^o\right)=143.1cm \end{gathered}$$

Magnitude ${\stackrel{\rightharpoonup }{d}}_{3m}=\sqrt{{\left(92.1\right)}^2+{\left(143.1\right)}^2}=170.1cm$

Direction $\theta ={\tan }^{-1}\frac{143.1}{92.1}=57.2^o$above +x axis

Calculation For Female

Similarly , We sum the components of the two vector for the female

$$\begin{gathered} {\stackrel{\rightharpoonup }{d}}_{3fx}={\stackrel{\rightharpoonup }{d}}_{1fx}+{\stackrel{\rightharpoonup }{d}}_{2fx}=0+\left(86.0\times \cos {\left(28.0\right)}^o\right)=75.9cm \\ {\stackrel{\rightharpoonup }{d}}_{3fy}={\stackrel{\rightharpoonup }{d}}_{1fy}+{\stackrel{\rightharpoonup }{d}}_{2fy}=84.0+\left(86.0\times \sin {\left(28.0\right)}^o\right)=124.4cm \end{gathered}$$

Magnitude ${\stackrel{\rightharpoonup }{d}}_{3f}=\sqrt{{\left(75.9\right)}^2+{\left(124.4\right)}^2}=145.7cm$

Direction $\theta ={\tan }^{-1}\frac{124.4}{75.9}=58.6^o$above +x axis