Question

In Exercises \(93-96,\) find the distance between the point and the line given by the set of parametric equations. (1,5,-2)\(; \quad x=4 t-2, \quad y=3, \quad z=-t+1\)

Short answer

So, the distance from given point (1,5,-2) to the line is \( \frac{7}{\sqrt{17}} \)

Step-by-step solution

Extract information from parametric equations

The parametric equations of the line are \(x=4t-2\), \(y=3\), \(z=-t+1\). From here the direction vector of the line can be obtained. The direction vector is given by the coefficients of 't', so \(d = (4, 0, -1)\). A point on the line can be found by setting t=0, so \(P_{line} = (-2, 3, 1)\)

Compute Vector and Unit Direction Vector

Next, compute the vector \( v \) between \( P_{line} \) and the given point (1,5,-2), which is \( v = P_{line} - P_{given} = (-2-1, 3-5, 1-(-2)) = (-3,-2,3) \). Now, compute the unit direction vector \( d^u \) of the line. This can be done by dividing the direction vector d by its magnitude which is \( \sqrt{4^2 + 0^2 + (-1)^2} = \sqrt{17} \). So, \( d^u = (\frac{4}{\sqrt{17}}, 0, \frac{-1}{\sqrt{17}} ) \)

Compute the Distance

The distance from the point to the line is given by the absolute value of the dot product of the vector v and the unit direction vector \( d^u \). The dot product is given by \( v \cdot d^u = (-3*\frac{4}{\sqrt{17}} - 2*0 + 3*\frac{-1}{\sqrt{17}} ) = \frac{-7}{\sqrt{17}} \). The distance is therefore \( |-7/\sqrt{17}| = 7/\sqrt{17} \)