Let L be the line with parametric equations x 8 y 63t z 3

Let L be the line with parametric equations

x = 8
y = 6+3t
z = 32t

Find the vector equation for a line that passes through the point P=(2, 3, 10) and intersects L at a point that is distance 2 from the point Q=(8, 6, 3). Note that there are two possible correct answers. Use the square root symbol \'\' where needed to give an exact value for your answer.

[x]

[y]

[z]

=

[?]

[?]

[?]

+ t

[?]

[?]

[?]

Solution

we need to find the point at distance of 2 units from ( 8 , -6 , 3)

sqrt(2) = sqrt[(8 - 8)^2 + ( -6 +6 -3t)^2 + (3 - 3 +2t)^2 ]

= sqrt{ 9t^2 + 4t^2}

sqrt2 = t*sqrt(13)

t = sqrt(2/13)

Point A = ( 8 , -6+ 3sqrt(2/13) , 3 - 2sqrt(2/13) )

P = ( 2, 3, -10)

Let other point ( x, y , z)

direction vector = < 6 , -9 +3sqrt(2/13) 13- 2sqrt(2/13) >

Equation of line: x -2 = 6t

y - 3= -9t +3tsqrt(2/13)

z +10 = 13t -2tsqrt(2/13)