Line and Planes?

Question

Find the distance from the point P = (1, 5, -7), to the line given by the vector equation
X = tv, where

> X := `<,>`(x, y, z);
> v := `<,>`(-2, 1, 0);

Answer 1

Direction cosines of of the line given by vector is
(x+2), (y-1),(z+7)...................i

Direction cosines of the line joining the points (1, 5, -7), and (-2, 1, 0); will be{1-(-2)},(5-1),(-7-0)
or 3, 4,-7.........................ii
The dot product of i and ii must be equal to zero
3(x+2)+4(y-1) -7(z+7) =0
3x +4y-7z=-6+4+49
3x +4y-7z-47=0.........................iii
Now find the distance from the point (1,5,-7) to the line(iii)
p={3x1 +4*5 -7*(-7) -47}/sq rt {3^2+4^2+(-7)^2}
p={3+20+49-47}/sq rt(9+16+49)
p=25/sq rt(74)=2.91 unit approximately

Answer 2

try asking your teacher