I Vector equation perpendicular to two equations

[53Mark53]

How do I find the vector equation of the line which passes through (-3/2,-3/2,1/2) and is perpendicular to both x+1=y/3=-z and 2x+1=2y+1=z-5/2

I know how to do it using one equation but I am unsure about how to do it using two equations

Thanks

 

[ShayanJ]

Find tangent vector to those lines and take their cross product. Then find the equation of a line with a tangent equal to that cross product which passes from that point.

 

[53Mark53]

Find tangent vector to those lines and take their cross product. Then find the equation of a line with a tangent equal to that cross product which passes from that point.

Could you please show an example I can't find anything about tangent vectors in my book

Thanks

 

[ShayanJ]

For example for the line $x+1=\frac y 3=-z$: At first we should transform it to the parametric form, simply write $z=-t$, then we'll have $\frac y 3=t \Rightarrow y=3t$ and $x=t-1$. So each point on the line has coordinates of the form $(t-1,3t,-t)$ and the tangent vector to this line is simply the derivative of the latter expression w.r.t. t which is $(1,3,-1)$.

 

[53Mark53]

For example for the line $x+1=\frac y 3=-z$: At first we should transform it to the parametric form, simply write $z=-t$, then we'll have $\frac y 3=t \Rightarrow y=3t$ and $x=t-1$. So each point on the line has coordinates of the form $(t-1,3t,-t)$ and the tangent vector to this line is simply the derivative of the latter expression w.r.t. t which is $(1,3,-1)$.

My class has not learned anything about using derivatives with vectors is there another way to solve this without using derivatives?

thanks

 

[ShayanJ]

My class has not learned anything about using derivatives with vectors is there another way to solve this without using derivatives?

thanks

What textbook are you using?
I think its better for me to take a look at it to see what tools you have at hand.

 

[53Mark53]

What textbook are you using?
I think its better for me to take a look at it to see what tools you have at hand.

This is the example we had from class

 

[ShayanJ]

Can you write the equation of those two lines in the form $(x,y,z)=\vec a + \vec b t$? This is the form in which the equation of the line is given in your example from the class.

 

[53Mark53]

Can you write the equation of those two lines in the form $(x,y,z)=\vec a + \vec b t$? This is the form in which the equation of the line is given in your example from the class.

(x,y,z)=(-1/2,-1/2-5/2)+t(1/2,1/2,1)
and
(x,y,z)=(-1,0,0)+s(1,3,-1)

would I have to combine these?

 

[ShayanJ]

(x,y,z)=(-1/2,-1/2-5/2)+t(1/2,1/2,1)
and
(x,y,z)=(-1,0,0)+s(1,3,-1)

would I have to combine these?

Good. Now those constant vectors which are the coefficients of t and s are the tangent vectors to those lines. Now you can either take the vector product of those tangent vectors as the tangent vector for your line or consider a vector with unknown components and set its scalar product with both tangent vectors to zero and find the components and take it as the tangent vector for your line.

 

[53Mark53]

Good. Now those constant vectors which are the coefficients of t and s are the tangent vectors to those lines. Now you can either take the vector product of those tangent vectors as the tangent vector for your line or consider a vector with unknown components and set its scalar product with both tangent vectors to zero and find the components and take it as the tangent vector for your line.

Thanks I got it now