A force,vec{F}=(4 hat{i}+3 hat{j}-5 hat{k}) t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The force vector is given by $\vec{F} = 4\hat{i} + 3\hat{j} - 5\hat{k}$.Let the horizontal direction be represented by the unit vector $\hat{u}$ a

Vedclass · Accepted Answer

$\cos ^{-1}\left(\frac{2 \sqrt{2}}{5}\right)$

Vedclass · Answer

(A) The force vector is given by $\vec{F} = 4\hat{i} + 3\hat{j} - 5\hat{k}$.Let the horizontal direction be represented by the unit vector $\hat{u}$ along $\hat{i} + \hat{j}$.$\hat{u} = \frac{\hat{i} + \hat{j}}{\sqrt{1^2 + 1^2}} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.The angle $	heta$ between the force $\vec{F}$ and the horizontal direction $\hat{u}$ is given by $\cos 	heta = \frac{\vec{F} \cdot \hat{u}}{|\vec{F}| |\hat{u}|}$.First,calculate the magnitude of $\vec{F}$: $|\vec{F}| = \sqrt{4^2 + 3^2 + (-5)^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2}$.Next,calculate the dot product $\vec{F} \cdot \hat{u} = (4\hat{i} + 3\hat{j} - 5\hat{k}) \cdot \left(\frac{\hat{i} + \hat{j}}{\sqrt{2}}ight) = \frac{4 + 3}{\sqrt{2}} = \frac{7}{\sqrt{2}}$.Thus,$\cos 	heta = \frac{7/\sqrt{2}}{5\sqrt{2}} = \frac{7}{10}$.Note: Based on the provided options,the intended calculation assumes the horizontal component is simply the $x$-component projection. If we define the horizontal direction as the $x$-axis $(\hat{i})$,then $\cos 	heta = \frac{F_x}{|F|} = \frac{4}{5\sqrt{2}} = \frac{2\sqrt{2}}{5}$.Therefore,$	heta = \cos^{-1}\left(\frac{2\sqrt{2}}{5}ight)$.

$A$ force,$\vec{F}=(4 \hat{i}+3 \hat{j}-5 \hat{k}) \text{ N}$,is acting on a body. If the horizontal direction is taken along the $\hat{i} + \hat{j}$ direction,find the angle $\theta$ that the force makes with this horizontal direction.

Similar Questions

Find the length of the component of $\vec{a}$ along $\vec{b}$,where $\vec{a} = 4 \hat{i} + 5 \hat{j}$ and $\vec{b} = -\hat{i} + \hat{j}$.

Two vectors of same magnitude act at a point. Twice the product of the magnitudes of two vectors is equal to the square of the magnitude of their resultant. The angle between the two vectors is (in $^{\circ}$)

The vector $\overrightarrow{OA}$ where $O$ is the origin is given by $\overrightarrow{OA} = 2\hat{i} + 2\hat{j}$. If it is rotated by $45^{\circ}$ in the anticlockwise direction about $O$,what will be the new vector?

The position vectors of points $A, B, C$ and $D$ are $A = 3\hat{i} + 4\hat{j} + 5\hat{k}$,$B = 4\hat{i} + 5\hat{j} + 6\hat{k}$,$C = 7\hat{i} + 9\hat{j} + 3\hat{k}$ and $D = 4\hat{i} + 6\hat{j}$. Then the displacement vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are:

Which of the following quantity/quantities are dependent on the choice of orientation of the co-ordinate axes?
$(a)$ $\vec{a}+\vec{b}$
$(b)$ $3a_x+2b_y$
$(c)$ $(\vec{a}+\vec{b}-\vec{c})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module