The first type the vector multiplication is referred to as the period product. The period product of vectors \(\vec A\) and \(\vec B\) is the sum of the assets of each pair of components. The is,


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Notice that the result of the dot product is a scalar, no a vector; that is sometimes dubbed scalar multiplication because that this reason. The period product of 2 vectors is likewise equal to
where θ is the angle in between the two vectors. This gives the period product some amazing features:

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If two vectors point in the very same direction (θ = 0°), climate their period product is simply the product of your magnitudes. In details
If two vectors point in the same-ish direction (that is, if the angle between them is less than 90°), climate their dot product is positive since the cosine of an acute angle is positive. The dot product of two vectors at right 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. Higher than 90°), so that they suggest in opposite-ish directions, climate their period product is negative.


Find the period product of \((3\hat x+\hat y)\) and \((-\hat x+4\hat y)\). Carry out the 2 vectors point in the same-ish or opposite-ish directions?
multiply the like components and include them together: $$(3)(-1)+(1)(4)=-3+4=\fbox1$$ The period product is positive, for this reason the vectors allude in the same-ish direction (that is, the angle in between them is acute.)
The period product of the 2 vectors is $$(3)(0)+(1)(4)=4$$ however the dot product is additionally the product of the size of the 2 vectors times the cosine that the angle between them. Therefore $$4=|\hat y||3\hat x+4\hat y|\cos\theta$$ The size of \(\hat y\) is 1, and also the magnitude of \(3\hat x+4\hat y\) is \(\sqrt3^2+4^2=5\), and also so $$4=5\cos\theta \implies \cos\theta=4\over5$$ $$\implies \theta=\cos^-10.8=\fbox$36.9^\circ$$$
If they are perpendicular, their period product should be zero: $$\begineqnarray 0&=&(3\hat x+4\hat y)\cdot(4\hat x+a\hat y)\cr &=&12+4a\cr 4a&=&-12\cr a&=&\fbox$-3$\cr \endeqnarray $$ because of this \(3\hat x+4\hat y\) and \(4\hat x-3\hat y\) room perpendicular to every other. In general, if you have actually a vector \(a\hat x+b\hat y\) and also you need another vector i m sorry is perpendicular come it, you have the right to use \(b\hat x-a\hat y\).
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The vector \(\vec F=\hat x+\hat z\) points follow me the face diagonal, and also the vector \(\vec B=\hat x+\hat y+\hat z\) points follow me the body diagonal; hence the angle between these two vectors is the edge θ us seek. Native the an interpretation of the dot 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 period product between the 2 is \((\hat x+\hat z)\cdot(\hat x+\hat y+\hat z)=1(1)+0(1)+1(1)=2\). Hence $$\theta=\cos^-1\left(\frac2\sqrt2\sqrt3\right)=\cos^-10.8165=\fbox$35.2^\circ$$$


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In the demo below, you deserve to rotate the 2 arrows around and also see just how the period product the the two is calculated.