91影视

The given vectors are,

${\stackrel{鈫�}{\mathrm{d}}}_1=4.0\stackrel{铃�}{\mathrm{i}}+5.0\stackrel{铃�}{\mathrm{j}}-6.0\stackrel{铃�}{\mathrm{k}}$

${\stackrel{鈫�}{\mathrm{d}}}_2=-1.0\stackrel{铃�}{\mathrm{i}}+2.0\stackrel{铃�}{\mathrm{j}}+3.0\stackrel{铃�}{\mathrm{k}}$

${\stackrel{鈫�}{\mathrm{d}}}_3=4.0\stackrel{铃�}{\mathrm{i}}+3.0\stackrel{铃�}{\mathrm{j}}+2.0\stackrel{铃�}{\mathrm{k}}$

Vector addition and subtraction can be used to find the displacement and magnitude can be found using the formula. To find the angle between $\stackrel{\mathbf{鈫�}}{\mathbf{r}}$and positive z axis we can use a scalar product of two vectors. Unit vector along z axis is. The components of ${\stackrel{\mathbf{鈫�}}{\mathbf{d}}}_{\mathbf{1}}$can be found by using the property of the scalar product of two vectors.

Formula:

$\stackrel{鈫�}{\mathrm{r}}={\stackrel{鈫�}{\mathrm{d}}}_1+{\stackrel{鈫�}{\mathrm{d}}}_2+{\stackrel{鈫�}{\mathrm{d}}}_3............\left(\mathrm{i}\right)$

${\stackrel{鈫�}{\mathrm{d}}}_1.{\stackrel{鈫�}{\mathrm{d}}}_1={\mathrm{d}}_1{\mathrm{d}}_2\mathrm{肠辞蝉蠁}.............\left(\mathrm{ii}\right)$

From equation (i), we can calculate the displacement vector $\stackrel{鈫�}{r}$as follows.

$$\begin{gathered} \stackrel{鈫�}{r}=\left(4.0\stackrel{铃�}{i}+5.0\stackrel{铃�}{j}-6.0\stackrel{铃�}{k}\right)-\left(-1.0\stackrel{铃�}{i}+2.0\stackrel{铃�}{j}+3.0\stackrel{铃�}{k}\right)+\left(4.0\stackrel{铃�}{i}+3.0\stackrel{铃�}{j}-2.0\stackrel{铃�}{k}\right) \\ =4.0\stackrel{铃�}{i}+5.0\stackrel{铃�}{j}-6.0\stackrel{铃�}{k}+1.0\stackrel{铃�}{i}-2.0\stackrel{铃�}{j}-3.0\stackrel{铃�}{k}+4.0\stackrel{铃�}{i}+3.0\stackrel{铃�}{j}+2.0\stackrel{铃�}{k} \\ =9.0\stackrel{铃�}{i}+6.0\stackrel{铃�}{j}-7.0\stackrel{铃�}{k} \end{gathered}$$

Therefore, the displacement$\stackrel{鈫�}{r}$is$9.0\stackrel{铃�}{i}+6.0\stackrel{铃�}{j}-7.0\stackrel{铃�}{k}$.

The magnitude of r is calculated as,

$$\begin{gathered} r=\sqrt{{\left(9.0\right)}^2+{\left(6.0\right)}^2+{\left(-7.0\right)}^2} \\ =12.9\mathrm{m} \end{gathered}$$

The angle between$\stackrel{鈫�}{r}$and positive z axis is 胃

localid="1658466997778" $$\begin{gathered} \cos 胃=\frac{\stackrel{鈫�}{r}.\stackrel{铃�}{k}}{\left|\stackrel{鈫�}{r}\right|} \\ =\frac{-7.0}{12.0} \\ 胃={\cos }^{-1}\left(-0.543\right) \\ =123掳 \end{gathered}$$

The angle betweenlocalid="1658467048657" $\stackrel{鈫�}{r}$and positive z axis is $123掳$.

The component of ${\stackrel{鈫�}{d}}_1$along the direction of${\stackrel{鈫�}{d}}_1={\stackrel{鈫�}{d}}_{\left|\right|}$

$$\begin{gathered} {\stackrel{鈫�}{d}}_{鈻�}={\stackrel{鈫�}{d}}_1.\stackrel{铃�}{u} \\ =d_1\cos 蠒.............\left(\mathrm{iii}\right) \end{gathered}$$

$\stackrel{铃�}{u}$is a unit vector

$$\begin{gathered} {\stackrel{鈫�}{d}}_1.{\stackrel{鈫�}{d}}_2=d_1{d}_2\cos 蠒 \\ \cos 蠒=\frac{{\stackrel{鈫�}{d}}_1\stackrel{鈫�}{d_2}}{d_1{d}_2}.............\left(iv\right) \end{gathered}$$

Equation (iii) becomes as

$$\begin{gathered} \stackrel{鈫�}{d_{鈻�}}=d_1\cos 蠒 \\ =d_1\left(\frac{{\stackrel{鈫�}{d}}_1.{\stackrel{鈫�}{d}}_2}{d_1{d}_2}\right) \\ =\frac{{\stackrel{鈫�}{d}}_1.{\stackrel{鈫�}{d}}_2}{d_2} \end{gathered}$$

Now, calculate the dot product${\stackrel{鈫�}{d}}_1.{\stackrel{鈫�}{d}}_2$ as,

role="math" localid="1658467900456" $$\begin{gathered} {\stackrel{鈫�}{d}}_1.{\stackrel{鈫�}{d}}_2=d_{1x}{d}_{2x}+d_{1y}{d}_{2y}+d_{1z}+d_{2z} \\ =\left(4.0\right)\left(-1.0\right)+\left(5.0\right)\left(2.0\right)+\left(-6.0\right)\left(3.0\right) \end{gathered}$$

Substitute it in the above equation.

role="math" localid="1658468009605" $$\begin{gathered} \stackrel{鈫�}{d_{鈻�}}=\frac{\left(4.0\right)\left(-1.0\right)+\left(5.0\right)\left(2.0\right)+\left(-6.0\right)\left(3.0\right)}{\sqrt{{\left(-1.0\right)}^2\left(2.0\right)+{\left(3.0\right)}^2}} \\ =-3.2\mathrm{m} \end{gathered}$$

The component of ${\stackrel{鈫�}{d}}_1$in the perpendicular direction of$\stackrel{鈫�}{d_2}=\stackrel{鈫�}{d_{鈯�}}$

$$\begin{gathered} {\stackrel{鈫�}{d}}_1={\stackrel{鈫�}{d}}_{鈯�}+{\stackrel{鈫�}{d}}_{鈻�} \\ {\stackrel{鈫�}{d}}_{鈯�}={\stackrel{鈫�}{d}}_1-{\stackrel{鈫�}{d}}_{鈻�} \\ {\stackrel{鈫�}{d}}_{鈯�}^2={\stackrel{鈫�}{d}}_1^2-d_{鈻�} \\ {d}_{鈯�}=\sqrt{\left[{\left(4.0\right)}^2+{\left(5.0\right)}^2+{\left(-6.0\right)}^2-{\left(-3.2\right)}^2\right]} \\ =8.2m \end{gathered}$$

Therefore, the component of ${\stackrel{鈫�}{d}}_1$in the perpendicular direction of ${\stackrel{鈫�}{d}}_2$is 8.2 m