Question

The Cartesian equation of a line is \(\frac{x+3}{2} = \frac{y-5}{4} = \frac{z+6}{2}\). Find its vector equation.

Hint:

To convert from Cartesian to vector form, simply pick out the coordinates of the point (remembering to flip the signs, e.g., \(x+3\) means \(x_1 = -3\)) to form vector \(\vec{a}\), and the denominators to form the direction vector \(\vec{b}\).

Answer 1

Step 1: Understanding the Concept:
The Cartesian equation of a line in 3D space, given by \(\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}\), describes a line that passes through the point \((x_1, y_1, z_1)\) and has direction ratios \(\langle a, b, c \rangle\). The vector equation of a line is given by \(\vec{r} = \vec{a} + \lambda \vec{b}\), where \(\vec{a}\) is the position vector of a point on the line and \(\vec{b}\) is a vector parallel to the line. We need to convert from one form to the other.
Step 2: Key Formula or Approach:
1. Identify the point \((x_1, y_1, z_1)\) through which the line passes from the Cartesian equation. This gives the position vector \(\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}\).
2. Identify the direction ratios \(\langle a, b, c \rangle\) from the denominators of the Cartesian equation. This gives the parallel vector \(\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}\).
3. Write the vector equation in the form \(\vec{r} = \vec{a} + \lambda \vec{b}\), where \(\lambda\) is a scalar parameter.
Step 3: Detailed Explanation:
The given Cartesian equation is: \[ \frac{x+3}{2} = \frac{y-5}{4} = \frac{z+6}{2} \] We can rewrite this in the standard form \(\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}\): \[ \frac{x - (-3)}{2} = \frac{y - 5}{4} = \frac{z - (-6)}{2} \] 1. Identify the point on the line:
Comparing with the standard form, the line passes through the point \((x_1, y_1, z_1) = (-3, 5, -6)\).
The position vector of this point is \(\vec{a} = -3\hat{i} + 5\hat{j} - 6\hat{k}\).
2. Identify the direction vector:
The direction ratios are given by the denominators, so \(\langle a, b, c \rangle = \langle 2, 4, 2 \rangle\).
The vector parallel to the line is \(\vec{b} = 2\hat{i} + 4\hat{j} + 2\hat{k}\).
3. Write the vector equation:
The vector equation of the line is \(\vec{r} = \vec{a} + \lambda \vec{b}\).
\[ \vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda (2\hat{i} + 4\hat{j} + 2\hat{k}) \] We can also use a simpler direction vector by factoring out a common scalar. Here, we can factor out 2 from \(\vec{b}\): \(2(\hat{i} + 2\hat{j} + \hat{k})\). Since we are using a parameter \(\lambda\), any scalar multiple of the direction vector works. Let's call the new parameter \(\mu = 2\lambda\). An equally valid vector equation is: \[ \vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \mu (\hat{i} + 2\hat{j} + \hat{k}) \] Step 4: Final Answer:
The vector equation of the line is \(\vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda (2\hat{i} + 4\hat{j} + 2\hat{k})\).