Question

Find the line through (8, 1, −5) that intersects and is perpendicular to the line x = −1 + t, y = −3 + t, z = −1 + t. (HINT: If (x0, y0, z0) is the point of intersection, find its coordinates. Enter your answers as a comma-separated list of equations.)

Answer 1

Answer:

$x=-6t+8,y=-t+1,z=7t-5$

Step-by-step explanation:

A line passing through the point (8, 1, −5) and perpendicular to the line x = −1 + t, y = −3 + t, z = −1 + t.

The parallel vector of given line is < 1, 1, 1 >.

Let the point of intersection is  (-1+t, -3+t, -1+t).

The required line passing through the points (8, 1, −5) and  (-1+t, -3+t, -1+t). So, the vector of required line is

< 8-(-1+t), 1-(-3+t), -5-(-1+t) >

< 9-t, 4-t, -4-t >

The dot product of two perpendicular vectors is 0.

Since < 9-t, 4-t, -4-t > and < 1, 1, 1 > are perpendicular vectors, therefore the dot product of these vectors is 0.

$(9-t)(1)+(4-t)(1)+(-4-t)(1)=0$

$9-t+4-t-4-t=0$

$9-3t=0$

$t=3$

The value of t is 3. So, the point of intersection is

$(-1+3, -3+3, -1+3)=(2,0,2)$

If a line passes through two points then the equation of line is

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

The line passes through two points (8, 1, −5) and (2,0,2). So, the equation of line is

$\frac{x-8}{2-8}=\frac{y-1}{0-1}=\frac{z+5}{2+5}$

Let

$\frac{x-8}{-6}=\frac{y-1}{-1}=\frac{z+5}{7}=t$

$x=-6t+8,y=-t+1,z=7t-5$

Therefore the parametric equations of the required line are $x=-6t+8,y=-t+1,z=7t-5$.