Question

Find the line through

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

Answer:

$(\frac{1}{2},\frac{-1}{2},\frac{1}{2})$

Step-by-step 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

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

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

$\Rightarrow -3+2t=0$

$2t=3$

$t=\frac{3}{2}$

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

$x=-1+\frac{3}{2}=\frac{1}{2}$

$y=-2+\frac{3}{2}=\frac{-1}{2}$

And $z=-1+\frac{3}{2}=\frac{1}{2}$