5. Vectors
c. Scalar Multiplication
1. Algebraic Definition
The scalar product of a scalar (real number), \(c\), and a vector, \(\vec v=\left\langle v_1,v_2\right\rangle\), is the vector whose components are the components of the original vector each multiplied by the scalar: \[ c\,\vec v=\left\langle c\,v_1,c\,v_2\right\rangle \]
If a particle moves \(2\) units east and \(3\) units north, then the displacement vector to describe this motion is \(\vec u=\left\langle2,3\right\rangle\). If this motion is repeated \(4\) times, then the total motion is \(4\vec u=4\left\langle2,3\right\rangle=\left\langle8,12\right\rangle\). That is \(8\) units east and \(12\) units north.
Compute \(-\dfrac{1}{2}\left\langle4,-5\right\rangle\).
\( -\dfrac{1}{2}\left\langle4,-5\right\rangle =\left\langle-2,\dfrac{5}{2}\right\rangle \)
\[ -\dfrac{1}{2}\left\langle4,-5\right\rangle =\left\langle-\dfrac{1}{2}\cdot4,-\dfrac{1}{2}\cdot(-5)\right\rangle =\left\langle-2,\dfrac{5}{2}\right\rangle \]
Recall that we previously defined the direction of a vector \(\vec v=\left\langle v_1,v_2\right\rangle\) to be the unit vector \[ \hat v =\left\langle\dfrac{v_1}{|\vec v|},\dfrac{v_2}{|\vec v|}\right\rangle \] and used the shorthand \[ \hat v=\dfrac{\vec v}{|\vec v|} \] for this vector. We now recognize that this notation is justified because it is the scalar multiplication of \(\vec v\) by the scalar \(\dfrac{1}{|\vec v|}\). Further, this equation can be solved for \(\vec v\) to give:
Every vector is the scalar product of its magnitude and its direction: \[ \vec v=|\vec v|\,\hat v \]
Find the vector \(\vec a\) of length \(4\) in the same direction as \(\vec b=\left\langle3,-4\right\rangle\).
\(\vec a =\left\langle\dfrac{12}{5},\dfrac{-16}{5}\right\rangle\)
The magnitude of \(\vec b=\left\langle3,-4\right\rangle\) is: \[ |\vec b|=\sqrt{9+16}=5 \] So the direction of \(\vec b\) (and \(\vec a\)) is the unit vector: \[ \hat a=\hat b=\dfrac{\vec b}{|\vec b|} =\left\langle\dfrac{3}{5},\dfrac{-4}{5}\right\rangle \] We want \(\vec a\) to have length of \(|\vec a|=4\). So \(\vec a\) is the product of its magnitude and its direction: \[ \vec a=|\vec a|\hat a =4\left\langle\dfrac{3}{5},\dfrac{-4}{5}\right\rangle =\left\langle\dfrac{12}{5},\dfrac{-16}{5}\right\rangle \]
We check the length of \(\vec a=\left\langle\dfrac{12}{5},\dfrac{-16}{5}\right\rangle\): \[ |\vec a|=\sqrt{\left(\dfrac{12}{5}\right)^2+\left(\dfrac{-16}{5}\right)^2} =\sqrt{\dfrac{144+256}{25}}=4 \]
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