Question 1 The line L passes through the points (1, 2, 3) and (3, 2, 1). Which one of the following equations is not a vector equation for the line? \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 3 \ 2 \ 1 \end{pmatrix} + s \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 3 \ 2 \ 1 \end{pmatrix} + s \begin{pmatrix} -2 \ 0 \ 2 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} + s \begin{pmatrix} 2 \ 0 \ -2 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} + s \begin{pmatrix} 2 \ 1 \ -2 \end{pmatrix}
Added by Rebecca P.
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Step 1
The slope is given by the formula: m = (y2 - y1) / (x2 - x1) Using the points (1, -1) and (1, 0), we have: m = (0 - (-1)) / (1 - 1) m = 1 / 0 Since the denominator is 0, the slope is undefined. This means that the line is vertical and parallel to the Show more…
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