Question

Find the line through

(7, 1, −6)
that intersects and is perpendicular to the line
x = −1 + t, y = −2 + 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: The answer is
`x=-1+2i,~y=-2+2j,~z=-1+2k.`

Step-by-step explanation: The vector equation of the line can be written as

`(x,y,z)=(-1,-2,-1)+t(i+j+k).`

The general equation of the plane that is perpendicular to this line will be

`x+y+z=c.`

Since it contains the point (7,1,-6), so

`7+1-6=c\\\\\Rightarrow c=2.`

Therefore, the plane x+y+z=2 contains the point (7,1,-6) and is perpendicular to the line.

Now, we will substitute the parametric equations of the line into the equation of the plane as follows

`x+y+z=2\\\\\Rightarrow (-1+t)+(-2+t)+(-1+t)=2\\\\\Rightarrow 3t=6\\\\\Rightarrow t=2.`

So, the parametric equation of the new line is

`(x,y,z)=(-1,-2,-1)+2(i+j+k)\\\\\Rightarrow x=-1+2i,~y=-2+2j,~z=-1+2k.`

Answer 2

`((1)/(2),(-1)/(2),(1)/(2))`

Explanation:

We have been given the intersection coordinates:

(7,1,-6) and perpendicular to the line x=-1+t ,y=-2+t and z=-1+t

From the condition of perpendicularity we know:

`(7,1,-6)(x,y,z)`

`\Rightarrow 7x+y-6z=0`

Now, we have given x=-1+t , y=-2+t and z= -1+t

`7(-1+t)+(-2+t)-6(-1+t)=0`

`-7+7t-2+t+6-6t=0`

`\Rightarrow -3+2t=0`

`2t=3`

`t=(3)/(2)`

Now, substituting t in the given coordinates x,y and z we get:

`x=-1+(3)/(2)=(1)/(2)`

`y=-2+(3)/(2)=(-1)/(2)`

And
`z=-1+(3)/(2)=(1)/(2)`