Find a parametrization for the line segment joining the points.
-(4, 0, 3), (0, 4, 0)
A) $x = - 3 t , y = - 3 t + 4 , z = - 2 t , 0 \leq t \leq 1$
B) $x = - 3 t + 4 , y = - 3 t , z = - 2 t + 3,0 \leq t \leq 1$
C) $x = 4 t , y = - 4 t + 4 , z = 3 t , 0 \leq t \leq 1$
D) $x = - 4 t + 4 , y = 4 t , z = - 3 t + 3,0 \leq t \leq 1$

Question

Find a parametrization for the line segment joining the points.
-(4, 0, 3), (0, 4, 0) 
A) $x = - 3 t , y = - 3 t + 4 , z = - 2 t , 0 \leq t \leq 1$
B) $x = - 3 t + 4 , y = - 3 t , z = - 2 t + 3,0 \leq t \leq 1$
C) $x = 4 t , y = - 4 t + 4 , z = 3 t , 0 \leq t \leq 1$
D) $x = - 4 t + 4 , y = 4 t , z = - 3 t + 3,0 \leq t \leq 1$

Quizplus · Accepted Answer

To parametrize the line segment between two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, we use the formula $P(t) = A + t(B - A)$, where $0 \leq t \leq 1$. For points $A(4, 0, 3)$ and $B(0, 4, 0)$, the parametrization becomes $x = 4 - 4t$, $y = 0 + 4t$, and $z = 3 - 3t$, which matches option D.

Find a Parametrization for the Line Segment Joining the Points x=−3t,y=−3t+4,z=−2t,0≤t≤1x = - 3 t , y = - 3 t + 4 , z = - 2 t , 0 \leq t \leq 1x=−3t,y=−3t+4,z=−2t,0≤t≤1

Find a Parametrization for the Line Segment Joining the Points $x = - 3 t , y = - 3 t + 4 , z = - 2 t , 0 \leq t \leq 1$