Question

Let P be a point not on the line L that passes through the points Q and R. Show that the distance d from the point P to the line L is Use the formula in part (a) to find the distance from the point P(1, 1, 1) to the line through Q(0, 7, 8) and R(−1, 5, 6).

Answer 1

Therefore, the distance from the point P(1, 1, 1) to the line through QR is 4.068.

Explanation:

We have the point P(1, 1, 1) to the line through Q(0, 7, 8) and R(−1, 5, 6).

Equation for line through PQ is

r= (0·i +7j+8k)+t ((-1-0)i +(5-7)j+ (6-8)k)

r=(7j+8k)+ t(-i-2j-2k)

We calculate the point T on line through RQ

T=\left[\begin{array}{ccc}T1\\T2\\T3\end{array}\right] =

\left[\begin{array}{ccc}0\\7\\8\end{array}\right] + t ·\left[\begin{array}{ccc}-1\\-2\\-2\end{array}\right]

T=\left[\begin{array}{ccc}T1\\T2\\T3\end{array}\right] =\left[\begin{array}{ccc}-t\\7-2t\\8-2t\end{array}\right]

PT=T-P = \left[\begin{array}{ccc}-t\\7-2t\\8-2t\end{array}\right] - \left[\begin{array}{ccc}1\\1\\1\end{array}\right]

PT=\left[\begin{array}{ccc}-1-t\\6-2t\\7-2t\end{array}\right]

Since PT is perpendicular to given line, then

\left[\begin{array}{ccc}-1-t\\6-2t\\7-2t\end{array}\right] · \left[\begin{array}{ccc}-1\\-2\\-2\end{array}\right]=0

we get

1+t-12+4t-14+4t=0

t=25/9

we have

PT=\left[\begin{array}{ccc}-1-t\\6-2t\\7-2t\end{array}\right]

PT=\left[\begin{array}{ccc}-34/9\\4/9\\13/9\end{array}\right]

We get that the distance from the point P(1, 1, 1) to the line through QR is

PT=\sqrt{(-34/9)^2 + (4/9)^2 + (13/9)^2}

PT=4.068

Therefore, the distance from the point P(1, 1, 1) to the line through QR is 4.068.