Question

If the Cartesian equation of a line is \( \frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} \), then find its equation in vector form.

Hint:

To quickly convert from Cartesian to vector form, identify the point by taking the values subtracted from x, y, and z (remember to flip the sign). The direction vector is formed by the denominators. This is a direct and fast conversion useful in timed exams.

Answer 1

Step 1: Understanding the Concept:
The Cartesian equation and the vector equation are two ways to represent a line in three-dimensional space. The vector form requires a point on the line and a direction vector for the line. Both of these can be extracted directly from the given Cartesian form.
Step 2: Key Formula or Approach:
The standard Cartesian form of a line is:
\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \] This line passes through the point \( P(x_0, y_0, z_0) \) and has direction ratios \( \langle a, b, c \rangle \).
The corresponding vector form of the equation is:
\[ \vec{r} = \vec{p} + \lambda \vec{d} \] where \( \vec{p} \) is the position vector of the point P, and \( \vec{d} \) is the direction vector of the line.
So, \( \vec{p} = x_0\hat{i} + y_0\hat{j} + z_0\hat{k} \) and \( \vec{d} = a\hat{i} + b\hat{j} + c\hat{k} \).
Step 3: Detailed Explanation or Calculation:
The given Cartesian equation is:
\[ \frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} \] We can rewrite the 'y' term to match the standard form \( y - y_0 \):
\[ \frac{x-5}{3} = \frac{y-(-4)}{7} = \frac{z-6}{2} \] Comparing this with the standard form \( \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \), we can identify:
The point on the line is \( (x_0, y_0, z_0) = (5, -4, 6) \).
The direction ratios are \( \langle a, b, c \rangle = \langle 3, 7, 2 \rangle \).
Now, we construct the position vector of the point and the direction vector.
The position vector \( \vec{p} \) is:
\[ \vec{p} = 5\hat{i} - 4\hat{j} + 6\hat{k} \] The direction vector \( \vec{d} \) is:
\[ \vec{d} = 3\hat{i} + 7\hat{j} + 2\hat{k} \] Using the vector form \( \vec{r} = \vec{p} + \lambda \vec{d} \), we get:
\[ \vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k}) \] where \( \lambda \) is a scalar parameter.
Step 4: Final Answer:
The equation of the line in vector form is \( \vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k}) \).