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.

Are the following vectors collinear? \( \vec{u} = \left< 1,4,5 \right>,\vec{v} = \left< -2,-8,10 \right> \)

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What is the value of \(z\) to make the vectors collinear? \( \vec{u} = \left< 1,4,5 \right>,\vec{v} = \left< -2,-8,z \right> \)

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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**

Express the vector \( \vec{u} = \left< 120, -45, 5000 \right>\) as a
linear combination of the Standard Basis unit vectors \( \hat{\imath}, \hat{\jmath} \) and \( \hat{k} \).

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Can \(\vec{u} = \left<2,4,6 \right> \) be written as a linear combination of
\(\vec{a} = \left<1,6 \right> \) and \(\vec{b} = \left<-5,16 \right> \)?

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