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)${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{1}}\mathbf{+}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{\left(}\mathbf{b}\mathbf{\right)}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{1}}\mathbf{-}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{2}}\mathbf{-}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{1}}$?

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

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:

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

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

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.

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.

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.