The geometric definition of the period product says that the dot product in between two vectors $vca$ and $vcb$ is$$vca cdot vcb = |vca| |vcb| cos heta,$$where $ heta$ is the angle in between vectors $vca$ and $vcb$.Although this formula is nice for understanding the nature of the dot product,a formula because that the dot product in regards to vector components would do it much easier to calculate the period product in between two given vectors.

You are watching: Evaluate the dot product of the vectors in (figure 1).

As a an initial step, us look at the period product between standard unit vectors, i.e., the vectors $vci$, $vcj$, and also $vck$ of size one and parallel to the name: coordinates axes.

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The standard unit vectors in 3 dimensions. The traditional unit vectors in 3 dimensions, $vci$ (green), $vcj$ (blue), and $vck$ (red) are length one vectors that suggest parallel to the $x$-axis, $y$-axis, and also $z$-axis respectively. Relocating them through the mouse doesn"t readjust the vectors, together they always suggest toward the positive direction the their respective axis.

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Since the conventional unit vectors are orthogonal, we instantly conclude that the period product between a pair of distinct standard unit vectors is zero:eginalign* vci cdot vcj = vci cdot vck = vcj cdot vck=0.endalign*The dot product between a unit vector and also itself is also simple to compute. In this case, the edge is zero and also $cos heta=1$. Offered that the vectors are every one of length one, the dot products areeginalign* vci cdot vci = vcj cdot vcj = vck cdot vck=1.endalign*

The 2nd step is to calculate the period product in between two three-dimensional vectorseginalign* vca &= (a_1,a_2,a_3) = a_1vci + a_2vcj+a_3vck\ vcb &= (b_1,b_2,b_3) = b_1vci + b_2vcj+b_3vck.endalign*To execute this, we merely assert the for any type of three vectors $vca$, $vcb$, and $vcc$, and any scalar $lambda$,eginalign* (lambdavca) cdot vcb &= lambda(vcacdotvcb) = vca cdot (lambdavcb)\ (vca+vcb) cdot vcc &= vca cdot vcc + vcbcdot vcc.endalign*(These properties mean that the dot product is linear.)

Given this properties and the fact that the dot product is commutative, we can expand the dot product $vca cdot vcb$ in regards to components,eginalign* vca cdot vcb &= (a_1vci + a_2vcj+a_3vck) cdot (b_1vci + b_2vcj+b_3vck) \ &= a_1b_1 vci cdot vci + a_2b_2vcjcdotvcj + a_3b_3vckcdotvck \ &quad + (a_1b_2+a_2b_1)vcicdotvcj + (a_1b_3+a_3b_1)vcicdotvck \ &quad + (a_2b_3+a_3b_2)vcjcdot vck.endalign*Since we recognize the dot product of unit vectors, we can simplify the period product formula toegingather vca cdot vcb = a_1b_1+a_2b_2+a_3b_3. labeldot_product_formula_3d ag1endgather

Equation eqrefdot_product_formula_3d provides it basic to calculate the period product of two three-dimensional vectors, $vca, vcb in R^3$.The equivalent equation because that vectors in the plane, $vca, vcb in R^2$,is also simpler. Giveneginalign* vca &= (a_1,a_2) = a_1vci + a_2vcj\ vcb &= (b_1,b_2) = b_1vci + b_2vcj,endalign*we deserve to use the very same formula, however with $a_3=b_3=0$,egingather vca cdot vcb = a_1b_1+a_2b_2 labeldot_product_formula_2d ag2.endgather

Armed through equations eqrefdot_product_formula_3d and eqrefdot_product_formula_2d, you have the right to make brief work that calculating period products, as presented in this examples.