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} = ) blank and ( p_0 = langle 1, -5, -3
angle ).

Explanation:

Step1: Recall parametric line formula

The parametric equations of a line in 3D space 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: Compare with given equations

Given \( x = 1 - t=1+(- 1)t \), \( y=-5 - 2t=-5+(-2)t \), \( z = - 3-5t=-3+(-5)t \).
Comparing with the general form, we have \( a=-1 \), \( b = - 2 \), \( c=-5 \) for the direction vector \( \mathbf{v} \), and \( x_0 = 1 \), \( y_0=-5 \), \( z_0=-3 \) for the point \( P_0 \).

Answer:

\(\langle - 1,-2,-5
angle\)