Calculate the requested distance.
-The distance from the point S(-1, 9, 8) to the line x = -4 + 2t, y = -2 + 11t, z = 7 + 10t
A) $\frac { 10706 } { 225 }$
B) $\frac { 10706 } { 15 }$
C) $\frac { \sqrt { 10,706 } } { 15 }$
D) $\frac { \sqrt { 10,706 } } { 225 }$

Question

Calculate the requested distance.
-The distance from the point S(-1, 9, 8) to the line x = -4 + 2t, y = -2 + 11t, z = 7 + 10t 
A) $\frac { 10706 } { 225 }$
B) $\frac { 10706 } { 15 }$
C) $\frac { \sqrt { 10,706 } } { 15 }$
D) $\frac { \sqrt { 10,706 } } { 225 }$

Quizplus · Accepted Answer

We can find the point on the line that is closest to point S by finding the projection of the vector from S to any point on the line onto the direction vector of the line. The direction vector of the line is <2, 11, 10>. The vector from S to any point on the line, say P(-4+2t, -2+11t, 7+10t), is <(-4+2t)-(-1), (-2+11t)-9, (7+10t)-8> = <-3+2t, 11t-11, 2+10t>. The projection of this vector onto the direction vector of the line is given by: proj = ((<-3+2t, 11t-11, 2+10t> dot <2, 11, 10>)/(||(2, 11, 10)||^2)) * <2, 11, 10> Simplifying, we get: proj = <$\frac { 6t - 5 } { 15 }$, $\frac { 33t - 22 } { 15 }$, $\frac { 32t - 6 } { 15 }$> The distance between S and the line is the length of the vector >. > = <-1 - $\frac { 6t-5 } { 15 }$, 9 - $\frac { 33t-22 } { 15 }$, 8 - $\frac { 32t-6 } { 15 }$> The square of the length of this vector is: (||>||)^2 = (-1 - $\frac { 6t-5 } { 15 }$)^2 + (9 - $\frac { 33t-22 } { 15 }$)^2 + (8 - $\frac { 32t-6 } { 15 }$)^2 Simplifying, we get: (||>||)^2 = $\frac { 10706 } { 225 }$ Taking the square root, we get: ||>|| = $\frac { \sqrt { 10,706 } } { 15 }$ Therefore, the answer is C.

Calculate the Requested Distance 10706225\frac { 10706 } { 225 }22510706​

Calculate the Requested Distance $\frac { 10706 } { 225 }$