Answer:
1) Angle with x-axis = 42.03 degrees
2) Angle with y-axis =68.2 degrees
3) Angle with z-axis = 56.14 degrees
Explanation:
given any vector [tex]\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}[/tex]
and any x axis the angle between them is given by
[tex]\theta_x =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{i}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{x\cdot i}{\sqrt{x^2+y^2+z^2}} )[/tex]
Applying values we get
[tex]\theta_x=cos^{-1}(\frac{160}{\sqrt{160^2+80^2+120^2}} )=42.03^{o}[/tex]
Angle between the vector and y axis is given by
[tex]\theta_y =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{j}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_y=cos^{-1}(\frac{y\cdot j}{\sqrt{x^2+y^2+z^2}} )[/tex]
Applying values we get
[tex]\theta_x=cos^{-1}(\frac{80}{\sqrt{160^2+80^2+120^2}} )=68.2^{o}[/tex]
Similarly angle between z axis and the vector is given by
[tex]\theta_z =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{k}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{z\cdot k}{\sqrt{x^2+y^2+z^2}} )[/tex]
Applying values we get
[tex]\theta_z=cos^{-1}(\frac{120}{\sqrt{160^2+80^2+120^2}} )=56.145^{o}[/tex]
The angles made by F will be "42.03°", "68.2°" and "56.14°".
According to the question,
Force, F = 160 i + 80 j + 120 kN
Let any vector,
[tex]\vec r = x \hat i + y \hat j+ z \hat k[/tex]
The angle between x-axis be:
[tex]\Theta_x[/tex] = Cos⁻¹ ([tex]\frac{\vec r. \hat i}{\sqrt{x^2 +y^2 + z^2} }[/tex])
= Cos⁻¹ ([tex]\frac{x.i}{\sqrt{x^2+y^2+z^2} }[/tex])
By substituting the values,
= Cos⁻¹ ([tex]\frac{160}{\sqrt{160^2+80^2+120^2} }[/tex])
= 42.03°
and,
The angle between y-axis be:
[tex]\Theta_y[/tex] = Cos⁻¹ ([tex]\frac{\vec r. \hat j}{\sqrt{x^2 +y^2 + z^2} }[/tex])
= Cos⁻¹ ([tex]\frac{y.i}{\sqrt{x^2+y^2+z^2} }[/tex])
By substituting the values,
= Cos⁻¹ ([tex]\frac{80}{\sqrt{160^2+80^2+120^2} }[/tex])
= 68.2°
and,
The angle between z-axis be:
[tex]\Theta_z[/tex] = Cos⁻¹ ([tex]\frac{\vec r. \hat k}{\sqrt{x^2 +y^2 + z^2} }[/tex])
= Cos⁻¹ ([tex]\frac{z.k}{\sqrt{x^2+y^2+z^2} }[/tex])
By substituting the values,
= Cos⁻¹ ([tex]\frac{120}{\sqrt{160^2+80^2+120^2} }[/tex])
= 56.145°
Thus the above approach is correct.
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