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Determine whether the vectors lie in a plane, and if so, whether they lie on a line. (a) v1=(1,2,1),v2=(7,1,3),v3=(14,1,4)\mathbf{v}_{1}=(1,2,1), \mathbf{v}_{2}=(-7,1,3), \mathbf{v}_{3}=(-14,-1,-4) (b) v1=(2,1,1),v2=(1,3,4),v3=(1,1,0)\mathbf{v}_{1}=(2,1,-1), \mathbf{v}_{2}=(1,3,4), \mathbf{v}_{3}=(1,1,0) (c) v1=(6,4,8),v2=(3,2,4),v3=(9,6,12)\mathbf{v}_{1}=(6,-4,8), \mathbf{v}_{2}=(-3,2,-4), \mathbf{v}_{3}=(9,-6,-12).

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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 v1,v2,v3R3\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\in \mathbb{R}^3 and A=[v1 v2 v3]A=[\mathbf{v}_1\ \mathbf{v}_2\ \mathbf{v}_3]. Then those vectors lie in a plane if rank(A)=2(A)=2. They lie on a line if rank(A)=1(A)=1 and they span R3\mathbb{R}^3 if rank (A)=3(A)=3. [15pt] a)\text{\color{#4257b2}a)} The matrix AA is given by:

[1714211134]\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:

[17140152701010]\begin{bmatrix} 1&-7&-14\\0&15&27\\0&10&10 \end{bmatrix}

Multiply the second row by 13\frac{1}{3} and the third by 110\frac{1}{10}:

[1714059011]\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:

[170040011]\begin{bmatrix} 1&7&0\\0&-4&0\\0&1&1 \end{bmatrix}

Multiply the second row by 14-\frac{1}{4}:

[170010011]\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:

[100010001]\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}

Therefore, rank(A)=3(A)=3 and thus the given vectors do not lie in a plane, they span R3\mathbb{R}^3

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