Question

Vector $\overrightarrow{\mathbf{A}}$has a negativexcomponent 3.00 unitsinlength and a positiveycomponent 2.00unitsin length.(a)Determine an expression for $\overrightarrow{\mathbf{A}}$in unit-vector notation.(b)Determine the magnitude and direction of $\overrightarrow{\mathbf{A}}$.(c)What vector $\overrightarrow{\mathbf{B}}$when added to $\overrightarrow{\mathbf{A}}$gives a resultant vector with noxcomponent and a negativeycomponent4.00 units in length?

Short answer

(a)$\stackrel{\rightharpoonup }{\mathrm{A}}=\left(-3\right)\overbrace{\mathrm{i}}+2\overbrace{\mathrm{j}}$

(b) 3.61 units and$146°$

(c)$\stackrel{\rightharpoonup }{\mathrm{B}}=3\hat{\mathrm{i}}-6\hat{\mathrm{j}}$

Step-by-step solution

Definition of vector

A vector is a quantity that has both a direction and a magnitude, and is used to determine the relative location of two points in space.

Step 2(a): Determine an expression for  in unit-vector notation A→

Unit- vector notation :

$\begin{array}{rcl}\stackrel{\rightharpoonup }{\mathrm{A}}& =& {\mathrm{A}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{A}}_{\mathrm{y}}\hat{\mathrm{j}}\\ \stackrel{\rightharpoonup }{\mathrm{A}}& =& \left(-3\right)\hat{\mathrm{i}}+2\hat{\mathrm{j}}\end{array}$

Hence, the expression is$\begin{array}{rcl}\stackrel{\rightharpoonup }{\mathrm{A}}& =& \left(-3\right)\hat{\mathrm{i}}+2\hat{\mathrm{j}}\end{array}$

Step 3(b): Determine the magnitude and direction of A→

The magnitudeof $\overrightarrow{\mathrm{A}}$:

$\begin{array}{rcl}& & \text{A =}\sqrt{{\text{A}}_{\text{x}}^{\text{2}}{\text{+ A}}_{\text{y}}^{\text{2}}}\\ & & \text{A =}\sqrt{{\text{( - 3)}}^{\text{2}}{\text{+2}}^{\text{2}}}\\ & =& \sqrt{13}\\ & =& 3.61\mathrm{units}\end{array}$

The direction of $\overrightarrow{\mathrm{A}}$:

$\begin{array}{rcl}\theta & =& {\tan }^{-1}\left(\frac{A_y}{A_x}\right)\\ & =& {\text{tan}}^{-1}\left(\frac{2}{-3}\right)\\ & =& -34°\end{array}$

According to thex andy components signs, the angle lies in the second quadrant.

The standard is to measure the angle counterclockwise from the +vex-axis

$\theta =180-34=146°$

Hence, the magnitude and the direction is 3.61 units and$146°$

Step 4(c): Determine the vector B→

$\begin{array}{rcl}\stackrel{\rightharpoonup }{\mathrm{R}}& =& \left(0\hat{\mathrm{i}}-4\hat{\mathrm{j}}\right)\\ \stackrel{\rightharpoonup }{\mathrm{R}}& =& \stackrel{\rightharpoonup }{\mathrm{B}}+\stackrel{\rightharpoonup }{\mathrm{A}}\\ \stackrel{\rightharpoonup }{\mathrm{B}}& =& \stackrel{\rightharpoonup }{\mathrm{R}}-\stackrel{\rightharpoonup }{\mathrm{A}}\end{array}$

Using the component method for adding vectors:

Calculate all thex components together.

Calculate all they components together.

role="math" localid="1663649637724" $\begin{array}{rcl}\stackrel{\rightharpoonup }{\mathrm{B}}& =& \left({\mathrm{R}}_{\mathrm{x}}-{\mathrm{A}}_{\mathrm{x}}\right)\hat{\mathrm{i}}+\left({\mathrm{R}}_{\mathrm{y}}-{\mathrm{A}}_{\mathrm{y}}\right)\hat{\mathrm{j}}\\ \stackrel{\rightharpoonup }{\mathrm{B}}& =& \left(0-(-3\right)\hat{\mathrm{i}}+\left(-4-2\right)\hat{\mathrm{j}}\\ \stackrel{\rightharpoonup }{\mathrm{B}}& =& 3\hat{\mathrm{i}}-6\hat{\mathrm{j}}\end{array}$

Hence, the vector is$\begin{array}{rcl}\stackrel{\rightharpoonup }{\mathrm{B}}& =& 3\hat{\mathrm{i}}-6\hat{\mathrm{j}}\end{array}$