Question

The two vectors shown in Fig. 3-21 lie in a xy plane. What are the signs

of the x and y components, respectively, of (a)${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{1}}\mathbf{+}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{\left(}\mathbf{b}\mathbf{\right)}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{1}}\mathbf{-}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{2}}\mathbf{-}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{1}}$?

Short answer

a. The x component is negative and the y component is positive for${\stackrel{\rightharpoonup }{d}}_1+{\stackrel{\rightharpoonup }{d}}_2$

b. The x component and y component of ${\stackrel{\rightharpoonup }{d}}_1-{\stackrel{\rightharpoonup }{d}}_2$is negative.

c. The x component and y component of ${\stackrel{\rightharpoonup }{d}}_2-{\stackrel{\rightharpoonup }{d}}_1$is positive.

Step-by-step solution

Given information

The two vectors are located in the XY plane as shown in the figure.

Vector addition and subtraction

The problem deals with the addition and subtraction of two vectors. Two vectors can be added or subtracted only when they are of the same type and nature. We can use the formula for the addition and subtraction of the two vectors.

Formula:

${\stackrel{\rightharpoonup }{d}}_1+\stackrel{\rightharpoonup }{d_2}=\stackrel{\rightharpoonup }{R}$

(a) To find the x and y component of d⇀1+d⇀2

In the given figure, the two vectors are in the xy plane.

a) X-Y component of$\stackrel{\rightharpoonup }{d_1}+\stackrel{\rightharpoonup }{d_2}$

The given figure A,$\stackrel{\rightharpoonup }{d_1}$has negative x-y components and$\stackrel{\rightharpoonup }{d_2}$has positive x-y components. When we draw the resultant$\stackrel{\rightharpoonup }{R}$, in the given figure, it would be present in the second quadrant.

$\stackrel{\rightharpoonup }{d_1}+\stackrel{\rightharpoonup }{d_2}=\stackrel{\rightharpoonup }{R}$

Hence according to the concept of addition of vector geometrically, the resultant has a negative x component and positive y component.

(b) To find the x and y components of d1⇀-d2⇀

For $\stackrel{\rightharpoonup }{d_1}-\stackrel{\rightharpoonup }{d_2}$, x-y components of $\stackrel{\rightharpoonup }{d_1}$and -$\stackrel{\rightharpoonup }{d_2}$are negative. When we draw the resultant $\stackrel{\rightharpoonup }{R}$by using the concept of subtraction of vectors geometrically, it would be present in the third quadrant as shown in figure B.

$\stackrel{\rightharpoonup }{d_1}-\stackrel{\rightharpoonup }{d_2}=\stackrel{\rightharpoonup }{R}$

Here,$-\stackrel{\rightharpoonup }{d_2}$has the same magnitude as$\stackrel{\rightharpoonup }{d_2}$but is oppositely directed.

Hence, $\stackrel{\rightharpoonup }{d_1}-\stackrel{\rightharpoonup }{d_2}$has negative x and y components.

(c) To find the x and y component of d2⇀-d1⇀

For,$\stackrel{\rightharpoonup }{d_2}-\stackrel{\rightharpoonup }{d_1}$ an x-y component of $-\stackrel{\rightharpoonup }{d_1}$and $\stackrel{\rightharpoonup }{d_2}$are positive. When we draw the resultant of $\stackrel{\rightharpoonup }{d_2}-\stackrel{\rightharpoonup }{d_1}$by using the concept of subtraction of vector geometrically, it can be located in the first quadrant as shown in figure C.

$\stackrel{\rightharpoonup }{d_2}-\stackrel{\rightharpoonup }{d_1}=\stackrel{\rightharpoonup }{R}$

Here, $-\stackrel{\rightharpoonup }{d_1}$has the same magnitude as $\stackrel{\rightharpoonup }{d_1}$, but is oppositely directed.

Hence, the x-y components of $\stackrel{\rightharpoonup }{d_2}-\stackrel{\rightharpoonup }{d_1}$are positive.