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 Techgenius May 27th, 2017 09:56 PM

Vector parallel to the line

Hi Guys! I was doing a vector math problem and I got stuck on how to find the vector parallel to the line.
The question is : Determine if the plane given by -x+2z=10 and the line given by r=(5,2-t,10+4t) are orthogonal, parallel or neither.

Now I have already found the normal vector as the first step, but also I have to find the vector that is parallel to the line so as that I can cross it with the normal vector to prove if it they are parallel to each other.
Thank you.

 skipjack May 28th, 2017 12:00 AM

Is there a point that lies in the given plane and on the given line?

 Country Boy May 28th, 2017 12:12 PM

Quote:
 Originally Posted by Techgenius (Post 571311) Hi Guys! I was doing a vector math problem and I got stuck on how to find the vector parallel to the line. The question is : Determine if the plane given by -x+2z=10 and the line given by r=(5,2-t,10+4t) are orthogonal, parallel or neither. Now I have already found the normal vector as the first step, but also I have to find the vector that is parallel to the line so as that I can cross it with the normal vector to prove if it they are parallel to each other. Thank you.
I presume that you have determined that a normal vector to the plane is <-1, 0, 2>. Is that what you got? Since your line is given by r = (5, 2 - t, 10 + 4t), taking t = 0, one point on that line is (5, 2, 10). Taking t = 1, another point is (5, 1, 14). The vector from (5, 2, 10) to (5, 1, 14) is <5 - 5, 1 - 2, 14 - 10> = <0, -1, 4> and that is a vector parallel to the line. Notice that <0, -1, 4> are the coefficients of t in "r = 5, 2 - t, 10 + 4t".

In general, a vector in the direction of the line \$\displaystyle x= x_0+ At\$, \$\displaystyle y= y_0+ Bt\$, \$\displaystyle z= z_0+ Ct\$ is <A, B, C>.

 Techgenius May 29th, 2017 05:55 AM

Quote:
 Originally Posted by Country Boy (Post 571345) I presume that you have determined that a normal vector to the plane is <-1, 0, 2>. Is that what you got? Since your line is given by r = (5, 2 - t, 10 + 4t), taking t = 0, one point on that line is (5, 2, 10). Taking t = 1, another point is (5, 1, 14). The vector from (5, 2, 10) to (5, 1, 14) is <5 - 5, 1 - 2, 14 - 10> = <0, -1, 4> and that is a vector parallel to the line. Notice that <0, -1, 4> are the coefficients of t in "r = 5, 2 - t, 10 + 4t". In general, a vector in the direction of the line \$\displaystyle x= x_0+ At\$, \$\displaystyle y= y_0+ Bt\$, \$\displaystyle z= z_0+ Ct\$ is .
Thank you very much Country Boy, you saved me in this. I now get the full picture. Thank you again.
#Slumerican

 zylo May 29th, 2017 11:25 AM