Question

A force, \( \vec{F} = (4\hat{i} + 3\hat{j} - 5\hat{k}) \) N is acting on a body making an angle \( \theta \) with the horizontal. Then the angle \( \theta \) is

• A. \( \cos^{-1} \left( \frac{2\sqrt{2}}{5} \right) \)
• B. \( \cos^{-1} \left( \frac{\sqrt{2}}{5} \right) \)
• C. \( \cos^{-1} \left( \frac{5\sqrt{2}}{9} \right) \)
• D. \( \cos^{-1} \left( \frac{3}{5\sqrt{2}} \right) \)

Hint:

The angle a vector makes with the horizontal (x-axis) can be found using the dot product of the vector with the unit vector along the x-axis, \( \hat{i} \). Use the formula \( \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \) and solve for \( \cos \theta \), then find \( \theta \) using the inverse cosine function. Remember to calculate the magnitude of the force vector correctly.

Answer 1

The Correct Option is A

The horizontal direction is usually represented by the x-axis.
So, we need to find the angle that the force vector \( \vec{F} = 4\hat{i} + 3\hat{j} - 5\hat{k} \) makes with the unit vector along the x-axis, which is \( \hat{i} \).
The magnitude of the force vector \( \vec{F} \) is: $$ |\vec{F}| = \sqrt{(4)^2 + (3)^2 + (-5)^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2} $$ The dot product of the force vector \( \vec{F} \) and the unit vector along the horizontal \( \hat{i} \) is: $$ \vec{F} \cdot \hat{i} = (4\hat{i} + 3\hat{j} - 5\hat{k}) \cdot (1\hat{i} + 0\hat{j} + 0\hat{k}) = 4(1) + 3(0) + (-5)(0) = 4 $$ The dot product of two vectors \( \vec{A} \) and \( \vec{B} \) is also given by \( \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \), where \( \theta \) is the angle between them.
In our case, \( \vec{A} = \vec{F} \) and \( \vec{B} = \hat{i} \).
The magnitude of \( \hat{i} \) is \( |\hat{i}| = 1 \).
$$ \vec{F} \cdot \hat{i} = |\vec{F}| |\hat{i}| \cos \theta $$ $$ 4 = (5\sqrt{2})(1) \cos \theta $$ $$ \cos \theta = \frac{4}{5\sqrt{2}} $$ To rationalize the denominator: $$ \cos \theta = \frac{4}{5\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}}{5 \times 2} = \frac{4\sqrt{2}}{10} = \frac{2\sqrt{2}}{5} $$ Therefore, the angle \( \theta \) is: $$ \theta = \cos^{-1} \left( \frac{2\sqrt{2}}{5} \right) $$