Find (a) “north cross west,” (b) “down dot south,” (c) “east cross up,” (d) “west dot west,” and (e) “south cross south.” Let each “vector” have unit magnitude

This problem is based on the product rule in which the vector product and scalar product are the two ways of multiplying vectors. To construct the scalar product, one multiplies the magnitude of the component of one vector by the magnitude of the component of the other vector. Similarly, When two vectors are multiplied by the sine of the angle between them, the magnitude of the vector product can be found.

Consider, eastward is $\hat{\mathrm{i}}$, northward is $\hat{\mathrm{j}}$, and upward is $\hat{\mathrm{k}}$respectively.

Using the fundamental product rule $\hat{\mathrm{i}},\hat{\mathrm{j}},\hat{\mathrm{k}}$can be written as

$\hat{\mathrm{i}}\times \hat{\mathrm{j}}=-\hat{\mathrm{j}}\times \mathrm{i}=\hat{\mathrm{k}}$ (i)

role="math" localid="1656308707958" $\hat{\mathrm{j}}\times \hat{\mathrm{k}}=-\hat{\mathrm{k}}\times \hat{\mathrm{j}}=\hat{\mathrm{i}}$ (ii) $\hat{\mathrm{k}}\times \hat{\mathrm{i}}=-\hat{\mathrm{i}}\times \hat{\mathrm{k}}=\hat{\mathrm{j}}$ (iii)

eastward is $\hat{\mathrm{i}}$, so west is $-\hat{\mathrm{i}}$therefore,

$\hat{\mathrm{j}}\times -\hat{\mathrm{i}}=\hat{\mathrm{k}}=\mathrm{up}$

Similarly,

$(-\hat{\mathrm{k}})·(-\hat{\mathrm{j}})=0$

$\hat{\mathrm{i}}\times \hat{\mathrm{k}}=-\hat{\mathrm{j}}\mathrm{south}\hat{\mathrm{i}}$

$(-\hat{\mathrm{i}})\cdot (-\hat{\mathrm{i}})=1$

$(-\hat{\mathrm{j}})\cdot (-\hat{\mathrm{j}})=1$