Question

suppose a line is described by the parametric equations ( x = 1 - t ), ( y = -5 - 2t ), ( z = -3 - 5t ). then the vector ( mathbf{v} ) and point ( p_0 ) that were used to define this line were ( mathbf{v} = langle -1, -2, -5
angle ) and ( p_0 = ) blank.

Explanation:

Step1: Recall parametric line formula

The parametric equations of a line in 3D are given by \( x = x_0 + at \), \( y = y_0 + bt \), \( z = z_0 + ct \), where \( \mathbf{v}=\langle a,b,c
angle \) is the direction vector and \( P_0=(x_0,y_0,z_0) \) is a point on the line.

Step2: Identify \( P_0 \) from given equations

Given \( x = 1 - t \), \( y = -5 - 2t \), \( z = -3 - 5t \). When \( t = 0 \), we get \( x = 1 \), \( y = -5 \), \( z = -3 \). So the point \( P_0 \) is \( (1, -5, -3) \).

Answer:

\( (1, -5, -3) \)