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one end of a long straight rod of length 4.0 m is located at (vec{r} = (12, 14, 10)) m and held in a position so that the angle between the rod and the +x axis is 55 degrees, the angle between the rod and the +y axis is 48 degrees, and the angle between the rod and the +z axis is 118 degrees. what is the location of the far end of the rod?

location of the far end of the rod = (langle quad , quad , quad
angle) m

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Explanation:

Step1: Recall the formula for direction cosines

The direction cosines \( \cos\alpha \), \( \cos\beta \), \( \cos\gamma \) are related to the angles \( \alpha \) (with x - axis), \( \beta \) (with y - axis), and \( \gamma \) (with z - axis) respectively. The unit vector \( \hat{\mathbf{u}} \) in the direction of the rod is given by \( \hat{\mathbf{u}}=\langle\cos\alpha,\cos\beta,\cos\gamma
angle \). Then the displacement vector \( \Delta\mathbf{r} \) of the rod is \( \Delta\mathbf{r}=L\hat{\mathbf{u}} \), where \( L = 4.0\space m \) is the length of the rod. The location of the far - end \( \mathbf{r}_f \) is given by \( \mathbf{r}_f=\mathbf{r}_i+\Delta\mathbf{r} \), where \( \mathbf{r}_i=\langle12,14,10
angle\space m \).

First, calculate the direction cosines:
\( \cos\alpha=\cos(55^{\circ})\approx0.5736 \)
\( \cos\beta=\cos(48^{\circ})\approx0.6691 \)
\( \cos\gamma=\cos(118^{\circ})=\cos(180^{\circ} - 62^{\circ})=-\cos(62^{\circ})\approx - 0.4695 \)

Step2: Calculate the displacement vector \( \Delta\mathbf{r} \)

The length of the rod \( L = 4.0\space m \). So,
\( \Delta x=L\cos\alpha=4.0\times0.5736 = 2.2944\space m \)
\( \Delta y=L\cos\beta=4.0\times0.6691 = 2.6764\space m \)
\( \Delta z=L\cos\gamma=4.0\times(- 0.4695)=- 1.878\space m \)

Step3: Calculate the location of the far - end

The initial position \( \mathbf{r}_i=\langle12,14,10
angle\space m \).
\( x_f=x_i+\Delta x=12 + 2.2944=14.2944\space m \)
\( y_f=y_i+\Delta y=14 + 2.6764=16.6764\space m \)
\( z_f=z_i+\Delta z=10-1.878 = 8.122\space m \)

Answer:

\( \langle14.3, 16.7, 8.12
angle \) (rounded to a reasonable number of significant figures. If we use more precise calculations: \( \cos(55^{\circ})\approx0.5735764364 \), \( \cos(48^{\circ})\approx0.6691306064 \), \( \cos(118^{\circ})\approx - 0.4694715628 \)

\( \Delta x = 4\times0.5735764364=2.294305746 \), \( x_f=12 + 2.294305746 = 14.29430575\approx14.3\)

\( \Delta y=4\times0.6691306064 = 2.676522426\), \( y_f=14 + 2.676522426=16.67652243\approx16.7\)

\( \Delta z=4\times(- 0.4694715628)=- 1.877886251\), \( z_f=10-1.877886251 = 8.122113749\approx8.12\))