Solve the problem.
-Find the point on the line of intersection of the planes x + y + z = 1 and 3x + 2y + z = 6 that is closest to the origin.
A) $\left( \frac { 7 } { 3 } , \frac { 1 } { 3 } , - \frac { 5 } { 3 } \right)$
B) $\left( \frac { 7 } { 3 } , - \frac { 1 } { 3 } , - \frac { 5 } { 3 } \right)$
C) $\left( - \frac { 7 } { 3 } , \frac { 1 } { 3 } , \frac { 5 } { 3 } \right)$
D) $\left( \frac { 7 } { 3 } , - \frac { 1 } { 3 } , \frac { 5 } { 3 } \right)$

Question

Solve the problem.
-Find the point on the line of intersection of the planes x + y + z = 1 and 3x + 2y + z = 6 that is closest to the origin. 
A) $\left( \frac { 7 } { 3 } , \frac { 1 } { 3 } , - \frac { 5 } { 3 } \right)$
B) $\left( \frac { 7 } { 3 } , - \frac { 1 } { 3 } , - \frac { 5 } { 3 } \right)$
C) $\left( - \frac { 7 } { 3 } , \frac { 1 } { 3 } , \frac { 5 } { 3 } \right)$
D) $\left( \frac { 7 } { 3 } , - \frac { 1 } { 3 } , \frac { 5 } { 3 } \right)$

Quizplus · Accepted Answer

The point closest to the origin on the line of intersection of two planes can be found by determining the direction vector of the line (cross product of the normals of the planes) and then finding the point on the line that is closest to the origin using projection formulas or optimization techniques. The correct calculation leads to the point $\left( \frac { 7 } { 3 } , \frac { 1 } { 3 } , - \frac { 5 } { 3 } \right)$.

Solve the Problem (73,13,−53)\left( \frac { 7 } { 3 } , \frac { 1 } { 3 } , - \frac { 5 } { 3 } \right)(37​,31​,−35​)

Solve the Problem $\left( \frac { 7 } { 3 } , \frac { 1 } { 3 } , - \frac { 5 } { 3 } \right)$