I need help on a homework problem. "Determine a vector equation for the line perpendicular to 4x-3y=17 and through point (-2,4). The answer in the book is given as [x,y]=[-2,4]+t[-4,3].Isn't the normal to the scalar equation [4,-3] and wouldn't this be the direction vector of the line we're looking for. I thought the answer should have been =[-2,4]+t[4,-3]. Where am I making a mistake?
To find the vector equation for a line perpendicular to a given line, we need to find the direction vector of the new line.
You correctly identified that the normal vector to the given line 4x - 3y = 17 is [4, -3]. However, this normal vector is not the direction vector we're looking for. Instead, the direction vector we need for the line perpendicular to 4x - 3y = 17 is the negative reciprocal of the direction vector [4, -3].
The negative reciprocal of [4, -3] is [-3, -4]. We flip the sign of each component and interchange their positions. So, the direction vector for the line perpendicular to 4x - 3y = 17 is [-3, -4].
Now, to find the vector equation for the line through the point (-2, 4) with direction vector [-3, -4], we can use the general equation for the line: [x, y] = [x₀, y₀] + t[a, b], where (x₀, y₀) is a point on the line, and (a, b) is the direction vector.
Plugging in the values, we have [x, y] = [-2, 4] + t[-3, -4].
So, the correct vector equation for the line perpendicular to 4x - 3y = 17 and through the point (-2, 4) is [x, y] = [-2, 4] + t[-3, -4].