4.3E: Exercises for Section 4.3

Exercise \(\PageIndex{1}\)

Find the vector equation for the line through \((−7, 6, 0)\) and \((−1, 1, 4)\). Then, find the parametric equations for this line.

Answer

\(\textbf{Vector Equation:}\) \[ \vec{r}(t) = \begin{bmatrix} -7 \\ 6 \\ 0 \end{bmatrix} + t \begin{bmatrix} 6 \\ -5 \\ 4 \end{bmatrix}, \quad t \in \mathbb{R}. \nonumber\]

\(\textbf{Parametric Equations:}\) \[ x = -7 + 6t, \quad y = 6 - 5t, \quad z = 4t, \quad t \in \mathbb{R}. \nonumber\]

Exercise \(\PageIndex{2}\)

Find parametric equations for the line through the point \((7, 7, 1)\) with a direction vector \(\vec{d}=\left[\begin{array}{c}1\\6\\2\end{array}\right]\).

Exercise \(\PageIndex{3}\)

Parametric equations of the line are \[\begin{aligned}x&=t+2 \\ y&=6-3t \\ x&=-t=6\end{aligned}\] Find a direction vector for the line and a point on the line.

Exercise \(\PageIndex{4}\)

Find the vector equation for the line through the two points \((−5, 5, 1),\: (2, 2, 4)\). Then, find the parametric equations.

Answer

\(\textbf{Vector Equation:}\) \[ \vec{r}(t) = \begin{bmatrix} -5 \\ 5 \\ 1 \end{bmatrix} + t \begin{bmatrix} 7 \\ -3 \\ 3 \end{bmatrix}, \quad t \in \mathbb{R}. \nonumber\]

\(\textbf{Parametric Equations:}\) \[ x = -5 + 7t, \quad y = 5 - 3t, \quad z = 1 + 3t, \quad t \in \mathbb{R}. \nonumber\]

Exercise \(\PageIndex{5}\)

The equation of a line in two dimensions is written as \(y = x−5\). Find parametric equations for this line.

Exercise \(\PageIndex{6}\)

Find parametric equations for the line through \((6, 5,−2)\) and \((5, 1, 2)\).

Exercise \(\PageIndex{7}\)

Find the vector equation and parametric equations for the line through the point \((−7, 10,−6)\) with a direction vector \(\vec{d}=\left[\begin{array}{c}1\\1\\3\end{array}\right]\).

Exercise \(\PageIndex{8}\)

Parametric equations of the line are \[\begin{aligned}x&=2t+2 \\ y&=5-4t \\ z&=-t-3\end{aligned}\nonumver\] Find a direction vector for the line and a point on the line, and write the vector equation of the line.

Answer

\(\textbf{Direction Vector:}\) The coefficients of \( t \) in the parametric equations give the direction vector: \[ \vec{d} = \begin{bmatrix} 2 \\ -4 \\ -1 \end{bmatrix}. \nonumber\]

\(\textbf{A Point on the Line:}\) Setting \( t = 0 \) in the parametric equations gives a point on the line: \[ P = (2, 5, -3). \nonumber\]

\(\textbf{Vector Equation:}\) The vector equation of the line is: \[ \vec{r}(t) = \begin{bmatrix} 2 \\ 5 \\ -3 \end{bmatrix} + t \begin{bmatrix} 2 \\ -4 \\ -1 \end{bmatrix}, \quad t \in \mathbb{R}. \nonumber\]

Exercise \(\PageIndex{9}\)

Find the vector equation and parametric equations for the line through the two points \((4, 10, 0),\: (1,−5,−6)\).

Exercise \(\PageIndex{10}\)

Find the point on the line segment from \(P = (−4, 7, 5)\) to \(Q = (2,−2,−3)\) which is \(\frac{1}{7}\) of the way from \(P\) to \(Q\).

Answer

The point that is \( \frac{1}{7} \) of the way from \( P = (-4,7,5) \) to \( Q = (2,-2,-3) \) is: \[ \left( \frac{-22}{7}, \frac{40}{7}, \frac{27}{7} \right). \nonumber\]

Exercise \(\PageIndex{11}\)

Suppose a triangle in \(\mathbb{R}^n\) has vertices at \(P_1,\: P_2,\) and \(P_3\). Consider the lines which are drawn from a vertex to the mid point of the opposite side. Show these three lines intersect in a point and find the coordinates of this point.