Problem 64 Hard Difficulty

(a) Find the point at which the given lines intersect:
$$ r = \langle 1, 1, 0 \rangle + t \langle 1, -1, 2 \rangle $$
$$ r = \langle 2, 0, 2 \rangle + s \langle -1, 1, 0 \rangle $$
(b) Find an equation of the plane that contains these lines.

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WZ

Wen Z.

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Video Transcript

the question they're asking to find the points at which the given lines intersect. The lines are given us our one equal to direction vector, one month zero plus team to direction vector one minus 12 and line are two is equal the direction vector 20 to plus S into direction vector minus +110. And the next question they're asking to find an equation on the plane that contains the above lines. So for the question A. Mhm. We can rearrange though. Equation of the first liners are one is equal to one plus T. I got Plus 1- T. Jacob. Yeah Plus two D. K. Cup. Mhm. And the Equation of flying to our two can be rewritten as mm two minus S. Icap plus S. Jacob plus to take up. Yeah. Therefore we can quit the coefficients of the first directions, I. J and K vectors. So uh in order to find the points of intersection of these lines. So therefore comparing the x coordinate, it is one plus two is equal to two minus S and one minus T is equal to S. And two is equal to two. So from equation three we get that T is equal to one. Putting this value of T is equal to one in the first equation. So we get one plus one equal to 2 -4. Therefore S is equal to zero. And putting this value of P N. S. That is S equal to zero, anti is equal to one in the rearranged equation. So If you put T is equal to one in In Equation one, R. 1 then we get the point of intersection equals do okay. Point of intersection will be one place Steven Dynasty in duty, one plus one, 1 -1 and two into 1. So this is equal to 202. And this is the point of intersection of the two lines. This is the required answer of the first question that this question A And the second question they are asking about the equation of the plane that contains these lines. So equation of the plane that contains there's lines is as we know the point of intersection. So The point is equal to 202. And in order to find the normal vector to the plane so this is equal to them cross productive the normal victims of buddha lines. So we want is equal to the coefficient of the direction vectors for each of the lines. So we won vector is equal to 1 -1, 2 and V two vector is equal to for the plane for the line to the direction vectors coefficients will be the normal victor. So we to vector is equal to -1, Therefore normal vector to the plane is equal to the cross product of these victims is equal to I J K gap With the Values V 1, -1, 2 and B 2 -1, 10. Yeah, so after computing the value of this, We kept it is equal to -2 icap -2, Jacob Plus zero K Cab. So therefore the normal victor to the plane and gap is equal to -2 -20. Therefore in order to find the equation of the plane. Mhm. Okay as austin which can be containing these lines is equal to the normal vector coefficients of the directions vectors. That is hygiene care respectively um minus two into x minus the point. Here is the point of intersection of the lines that is 202. So putting this value in the coordinates of the planes X, Y and Z. We get x minus two minus two. Why minus zero plus zero Into Zed -2 is equal to zero. So this is our equation of the plane so this is Equal to -2, x plus four minus two. White Is equal to zero. So this can be written equal to two X Plus two, y equal to four. So express why equal to 2? And this is the equation of the plane continuing the given lines are running are too and this is the answer of the question. B right