Linear Combinations
A vector is said to be a Linear Combination of another vector if there exists a non-zero scalar such that \( \vec{u} = k\vec{v} \) or \( \vec{u} = a\vec{v} + b\vec{w} \) etc. In general:
\(\vec{u} = \sum_{i=0}^n a_i\vec{v}_i = a_1\vec{v}_1 + a_2\vec{v}_2 + \cdots + a_n\vec{v}_n \)
Where \(a_1, a_2\), etc. are scalars and \(\vec{v}_1, \vec{v}_2\), etc. are vectors.
If \(\vec{u} = k\vec{v} \), \( \vec{u} \) is said to be Collinear with respect to \( \vec{v} \). In other words, \( \vec{u} \) is a scalar multiple of \( \vec{v} \), \(\vec{u} \) is parallel to \( \vec{v} \) and \( \vec{u} \) and \( \vec{v} \) point in the same direction.
\(\vec{u} \parallel \vec{v} \)
\(\vec{u} = 2.5 \vec{v}\)
If \(\vec{u} = a\vec{v} + b\vec{w}\), \(\vec{u}\) is said to be Coplanar with respect to \( \vec{v} \) and \( \vec{w} \). In other words, \( \vec{u} \), \(\vec{v}\) and \(\vec{w}\) all exist on the same plane in 3-D space.
\(\vec{w} = 2.5 \vec{v} + 2 \vec{u}\)
In 3-D, imagine that all three vectors can be drawn on a piece of paper.
We need to check if there exists a scalar \(k\) such that \(\vec{u} = k\vec{v}\). This equation needs to be true for all dimensions of the vector (the \(x\), \(y\), and \(z\)-directions):
\(\vec{u} = k\vec{v}\)
\(\left< 1,4,5 \right> = k\left< -2,-8,10 \right>\)
\(x\)-direction:
\( 1 = k(-2)\)
\( k = -\cfrac{1}{2}\)
\(y\)-direction:
\( 4 = k(-8)\)
\( k = -\cfrac{4}{8}=-\cfrac{1}{2}\)
\(z\)-direction:
\( 5 = k(10)\)
\(k = \cfrac{5}{10} = \cfrac{1}{2}\)
Since there is a contradiction in the value of \(k\), these vectors are not collinear.
Again, we want to find a scalar \(k\) such that \(\vec{u} = k\vec{v}\). From above, we know \(k = -\cfrac{1}{2} \) for the \(x\) and \(y\)- coordinates. Therefore:
\(5 = kz\)
\(5 = -\cfrac{1}{2}z\)
\(5(-2) = z\)
\( z = -10\)
The vectors \( \vec{u} = \left< 1,4,5 \right> \) and \(\vec{v} = \left< -2,-8,-10 \right> \) are collinear.
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Standard Basis Form
Recall the Standard Basis unit vectors that are in the direction of the the \(x\), \(y\) and \(z\)-axes:
\(\hat{\imath} = \left<1,0,0\right> \)
\(\hat{\jmath} = \left<0,1,0\right> \)
\(\hat{k} = \left<0,0,1\right> \)
We can express any vector in 3-D as a linear combination of these vectors. This is called Standard Basis Form
The vector has an \(x\)-component of \(120\). Since the unit vector \(\hat{\imath} \) has a magnitude of \(1\), we need to multiple by a scalar of \(120\). This will result in a vector that has a magnitude of \(120\) along the \(x\)-axis.
Next, we want to add a vector for the \(y\)-component. Multiply the unit vector \(\hat{\jmath} \) by \(-45\).
Lastly add a vector for the \(z\)-component by multiplying the unit vector \(\hat{k}\) by \(5000\).
We can express the vector in Standard Basis Form as such:
\(\vec{u} = \left< 120, -45, 5000 \right> = 120\hat{\imath} - 45\hat{\jmath} + 5000\hat{k}\)
Since vector \(\vec{u} \) has a \(z\)-component, it cannot be written as a linear combination of \(\vec{a}\) and \(\vec{b} \) as neither have \(z\)-components.