Question

Find a point \( P \) on the line \( \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9} \) such that its distance from point \( Q(2, 4, -1) \) is 7 units. Also, find the equation of the line joining \( P \) and \( Q \).

Hint:

When finding the point on a line at a specific distance from another point, use the distance formula and parametric equations. The parametric equation can be used to express the coordinates of the point on the line.

Answer 1

The line is defined by the equation: \[ \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9} = t. \] This yields the parametric equations: \[ x = -5 + t, \quad y = -3 + 4t, \quad z = 6 - 9t. \] Let \( P(x, y, z) \) be a point on this line. We are given that the distance between \( P \) and \( Q(2, 4, -1) \) is 7. The distance formula is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}. \] Applying this to \( P \) and \( Q \): \[ 7 = \sqrt{(x - 2)^2 + (y - 4)^2 + (z + 1)^2}. \] Substitute the parametric expressions for \( x, y, \) and \( z \) into this distance equation and solve for \( t \). Once \( t \) is found, substitute it back into the parametric equations to determine the coordinates of \( P \). Finally, find the equation of the line connecting \( P(x, y, z) \) and \( Q(2, 4, -1) \). The parametric form of a line through two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is: \[ \frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}. \] Substitute the coordinates of \( P \) and \( Q \) to obtain the equation of the line segment \( PQ \).