The first type the vector multiplication is dubbed the period product. The period product the vectors \(\vec A\) and \(\vec B\) is the sum of the commodities of each pair that components. That is,

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Notice the the result of the period product is a scalar, not a vector; that is sometimes called scalar multiplication for this reason. The period product of 2 vectors is likewise equal to

where θ is the angle between the two vectors. This offers the dot product some amazing features:

If two vectors suggest in the same direction (θ = 0°), then their dot product is just the product of their magnitudes. In specific

If 2 vectors point in the same-ish direction (that is, if the angle in between them is much less than 90°), climate their dot product is positive since the cosine of one acute edge is positive. The dot product of two vectors at appropriate angles come each various other is zero. Because that example, \(\hat x\cdot\hat y=0\). If the angle between two vectors is obtuse (i.e. Greater than 90°), so the they point in opposite-ish directions, then their period product is negative.

Find the dot product the \((3\hat x+\hat y)\) and also \((-\hat x+4\hat y)\). Carry out the 2 vectors suggest in the same-ish or opposite-ish directions?

multiply the like contents and include them together: $$(3)(-1)+(1)(4)=-3+4=\fbox1$$ The dot product is positive, so the vectors allude in the same-ish direction (that is, the angle between them is acute.)

The period product of the 2 vectors is $$(3)(0)+(1)(4)=4$$ however the period product is additionally the product of the size of the 2 vectors time the cosine of the angle in between them. Thus $$4=|\hat y||3\hat x+4\hat y|\cos\theta$$ The size of \(\hat y\) is 1, and the size of \(3\hat x+4\hat y\) is \(\sqrt3^2+4^2=5\), and so $$4=5\cos\theta \implies \cos\theta=4\over5$$ $$\implies \theta=\cos^-10.8=\fbox$36.9^\circ$$$

If they room perpendicular, their dot product need to be zero: $$\begineqnarray 0&=&(3\hat x+4\hat y)\cdot(4\hat x+a\hat y)\cr &=&12+4a\cr 4a&=&-12\cr a&=&\fbox$-3$\cr \endeqnarray $$ thus \(3\hat x+4\hat y\) and also \(4\hat x-3\hat y\) are perpendicular to every other. In general, if you have a vector \(a\hat x+b\hat y\) and you need one more vector i m sorry is perpendicular come it, you can use \(b\hat x-a\hat y\).

The vector \(\vec F=\hat x+\hat z\) points follow me the confront diagonal, and the vector \(\vec B=\hat x+\hat y+\hat z\) points follow me the human body diagonal; therefore the angle in between these two vectors is the angle θ us seek. From the an interpretation of the period product $$\vec F\cdot\vec B=|\vec F||\vec B|\cos\theta \implies \theta=\cos^-1\left(\frac\vec F\cdot\vec B\right)$$ The size of \(\vec F\) is \(|\vec F|=\sqrt1^2+0^2+1^2=\sqrt2\); the size of \(\vec B\) is \(\sqrt3\). The dot product in between the 2 is \((\hat x+\hat z)\cdot(\hat x+\hat y+\hat z)=1(1)+0(1)+1(1)=2\). Thus $$\theta=\cos^-1\left(\frac2\sqrt2\sqrt3\right)=\cos^-10.8165=\fbox$35.2^\circ$$$

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In the demo below, you can rotate the 2 arrows around and see how the period product the the two is calculated.