# Find the distance from the point (-2,3,-1) to the line x = 3+4t , y = 0+5t , z = 1+5t

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The distance between the given point A(-2,3,-1) and some point on the line B(3+4t, 5t, 1 + 5t) is the perpendicular distance.

Considering that the line `AB` is perpendicular to the direction vector of the given line, `bar v = <4,5,5>` , hence, the dot product `bar(AB)*bar v = 0` , such that:

`bar(AB)*bar v = -2*(3+4t) + 3*(5t) - 1*(1 + 5t) `

`bar(AB)*bar v = -6 - 8t + 15t - 1 - 5t`

`{(bar(AB)*bar v = 2t - 7),(bar(AB)*bar v = 0):} => 2t - 7 = 0 => t = 7/2`

Replacing `7/2` for `t` yields:

`B(3+4*(7/2),5*(7/2),1+5*(7/2)) = B(17,35/2,37/2)`

You may evaluate the distance `AB` using the distance formula, such that:

`AB = sqrt((x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2)`

`AB = sqrt((17 + 2)^2 + (35/2 - 3)^2 + (37/2 + 1)^2)`

`AB = sqrt(361 + 841/4 + 1521/4)`

`AB = sqrt(361 + 1181/2)`

`AB ~~ 30.84`

**Hence, evaluating the distance AB form the given point, to the given line, using distance formula, yields **`AB ~~ 30.84.`