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# Determine whether the vectors lie in a plane, and if so, whether they lie on a line. (a) $\mathbf{v}_{1}=(1,2,1), \mathbf{v}_{2}=(-7,1,3), \mathbf{v}_{3}=(-14,-1,-4)$ (b) $\mathbf{v}_{1}=(2,1,-1), \mathbf{v}_{2}=(1,3,4), \mathbf{v}_{3}=(1,1,0)$ (c) $\mathbf{v}_{1}=(6,-4,8), \mathbf{v}_{2}=(-3,2,-4), \mathbf{v}_{3}=(9,-6,-12)$.

Solution

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Notice that three vectors lie in the plane if one of them can be expressed as a linear combination of the other two. Three vectors lie in the line if two vectors are a scalar multiple of the third. Let $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\in \mathbb{R}^3$ and $A=[\mathbf{v}_1\ \mathbf{v}_2\ \mathbf{v}_3]$. Then those vectors lie in a plane if rank$(A)=2$. They lie on a line if rank$(A)=1$ and they span $\mathbb{R}^3$ if rank $(A)=3$. [15pt] $\text{\color{#4257b2}a)}$ The matrix $A$ is given by:

$\begin{bmatrix} 1&-7&-14\\2&1&-1\\1&3&-4 \end{bmatrix}$

Add -2 times the first row to the second and -1 times the first row to the third:

$\begin{bmatrix} 1&-7&-14\\0&15&27\\0&10&10 \end{bmatrix}$

Multiply the second row by $\frac{1}{3}$ and the third by $\frac{1}{10}$:

$\begin{bmatrix} 1&-7&-14\\0&5&9\\0&1&1 \end{bmatrix}$

Add -9 times the third row to the second and add 14 times the third row to the first:

$\begin{bmatrix} 1&7&0\\0&-4&0\\0&1&1 \end{bmatrix}$

Multiply the second row by $-\frac{1}{4}$:

$\begin{bmatrix} 1&7&0\\0&1&0\\0&1&1 \end{bmatrix}$

Add -7 times the second row to the first and -1 times the second row to the third:

$\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$

Therefore, rank$(A)=3$ and thus the given vectors do not lie in a plane, they span $\mathbb{R}^3$

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