2. Find a Cartesian equations of the line containing the point $$ A $$ and parallel to the vector $$ v $$. a) $$ A(-4,3,5), v=4 i+8 j-7 k $$ b) $$ A(-6,-4,5), v=5 i+8 j-4 k $$ c) $$ A(4,-5,9), v=6 i+8 j-14 k $$

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**(a) For $$ A(-4, 3, 5) $$ and $$ v = 4\mathbf{i} + 8\mathbf{j} -7\mathbf{k} $$:**

1. The parametric equations of the line are given by:
   $$
   x = -4 + 4t,\quad y = 3 + 8t,\quad z = 5 - 7t.
   $$

2. To write the symmetric (Cartesian) form, solve each for $$ t $$:
   $$
   t = \frac{x+4}{4},\quad t = \frac{y-3}{8},\quad t = \frac{5-z}{7}.
   $$
   Thus, the Cartesian equation is:
   $$
   \frac{x+4}{4} = \frac{y-3}{8} = \frac{5-z}{7}.
   $$

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**(b) For $$ A(-6, -4, 5) $$ and $$ v = 5\mathbf{i} + 8\mathbf{j} -4\mathbf{k} $$:**

1. The parametric equations of the line are:
   $$
   x = -6 + 5t,\quad y = -4 + 8t,\quad z = 5 - 4t.
   $$

2. Expressing $$ t $$ in each equation gives:
   $$
   t = \frac{x+6}{5},\quad t = \frac{y+4}{8},\quad t = \frac{5-z}{4}.
   $$
   Hence, the Cartesian equation is:
   $$
   \frac{x+6}{5} = \frac{y+4}{8} = \frac{5-z}{4}.
   $$

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**(c) For $$ A(4, -5, 9) $$ and $$ v = 6\mathbf{i} + 8\mathbf{j} -14\mathbf{k} $$:**

1. The parametric equations are:
   $$
   x = 4 + 6t,\quad y = -5 + 8t,\quad z = 9 - 14t.
   $$

2. Solving for $$ t $$ in each gives:
   $$
   t = \frac{x-4}{6},\quad t = \frac{y+5}{8},\quad t = \frac{9-z}{14}.
   $$
   The Cartesian equation is thus:
   $$
   \frac{x-4}{6} = \frac{y+5}{8} = \frac{9-z}{14}.
   $$
The Cartesian equations of the lines are: **(a)** $$ \frac{x+4}{4} = \frac{y-3}{8} = \frac{5-z}{7} $$ **(b)** $$ \frac{x+6}{5} = \frac{y+4}{8} = \frac{5-z}{4} $$ **(c)** $$ \frac{x-4}{6} = \frac{y+5}{8} = \frac{9-z}{14} $$

Find a Cartesian equations of the line containing the point and parallel to the vector .
a)
b)
c)

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