Find the intersection.
-x = 4 + 3t, y = 10 - 10t, z = 2 + 3t ; -2x + 2y - 2z = 9
A) $\left( \frac { 125 } { 32 } , \frac { 165 } { 16 } , \frac { 61 } { 32 } \right)$
B) $( 1,20 , - 1 )$
C) $\left( \frac { 131 } { 32 } , \frac { 155 } { 16 } , \frac { 63 } { 32 } \right)$
D) $( 7,0,5 )$

Question

Find the intersection.
-x = 4 + 3t, y = 10 - 10t, z = 2 + 3t ; -2x + 2y - 2z = 9 
A) $\left( \frac { 125 } { 32 } , \frac { 165 } { 16 } , \frac { 61 } { 32 } \right)$
B) $( 1,20 , - 1 )$
C) $\left( \frac { 131 } { 32 } , \frac { 155 } { 16 } , \frac { 63 } { 32 } \right)$
D) $( 7,0,5 )$

Quizplus · Accepted Answer

Substituting $x = 4 + 3t$, $y = 10 - 10t$, and $z = 2 + 3t$ into the equation $-2x + 2y - 2z = 9$ and solving for $t$ gives the specific values of $x$, $y$, and $z$ that satisfy both the parametric equations and the given equation, leading to the coordinates $\left( \frac { 125 } { 32 } , \frac { 165 } { 16 } , \frac { 61 } { 32 } \right)$.

Find the Intersection (12532,16516,6132)\left( \frac { 125 } { 32 } , \frac { 165 } { 16 } , \frac { 61 } { 32 } \right)(32125​,16165​,3261​)

Find the Intersection $\left( \frac { 125 } { 32 } , \frac { 165 } { 16 } , \frac { 61 } { 32 } \right)$