Consider the line ` in R^2 with normal vector n = [1, −5] and passing through the pointP = (−3, 4).(a) Write down equations in normal form and general form for this line.(b) Use the general form to find parametric equations for `, and then write down avector equation for ` as well.(c) Hence or otherwise write down a direction vector for `.ChatGPT

(a) Write down equations in normal form and general form for this line:

The normal form of a line is given by (r - P) · n = 0, where r is a position vector, P is a point on the line, and n is the normal vector.

Given P = (-3, 4) and n = [1, -5], we can write the normal form as:

(r - [-3, 4]) · [1, -5] = 0 (r + [3, -4]) · [1, -5] = 0 r · [1, -5] + [3, -4] · [1, -5] = 0 r · [1, -5] - 17 = 0

So, the equation of the line ` in normal form is r · [1, -5] - 17 = 0.

The general form of a line is given by Ax + By = C, where A and B are the components of the normal vector and C is a constant.

From the normal form equation, we can rewrite it as:

x - 5y = 17

So, the equation of the line ` in general form is x - 5y = 17.

(b) Use the general form to find parametric equations for `:

Rearranging the general form equation, we can express x and y in terms of a parameter t:

x = t y = (17 - t) / 5

So, the parametric equations for ` are:

x = t y = (17 - t) / 5

To write down a vector equation for `, we can combine the parametric equations into a single vector equation:

` = [t, (17 - t) / 5]

(c) Hence or otherwise write down a direction vector for `:

The direction vector of a line is perpendicular to the normal vector. Given the normal vector n = [1, -5], a direction vector d for the line ` can be found by rotating n by 90 degrees. This can be done by swapping the components of n and changing the sign of one of them.

So, a direction vector for ` is d = [5, 1].