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(a) Find parametric equations for the line throug…

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Problem 15 Medium Difficulty

(a) Find symmetric equations for the line that passes through the point $ (1, -5, 6) $ and is parallel to the vector $ \langle -1, 2, -3 \rangle $.
(b) Find the points in which the required line in part (a) intersects the coordinate planes.

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Video Transcript

in part a of this question, we are given a point and a vector and rest to find symmetric equations for the line. It passes through this point and is parallel to this vector. The point is one negative five six and the vector is negative. 12 negative three. Well, we know this metric equation of a line Passing through a point and parallel to a vector is given by the general Form X minus the first component of the point one over the first component of the vector Negative one. This is equal to wine minus the y component of the point. Negative five over the y component of the vector too. This is equal to Z minus the Z component of the 0.6 over the sea component of the vector Negative three. And so this is the symmetric equation. Mhm then in part B. Whereas to find the points in which the required line can part a intersects the coordinate planes, this is actually pretty easy to do using the symmetric form of the line, For example, we know the line intersects the X Y plane when Z is equal to zero. So if you set Z equal to zero. In these equations, we have X minus one over negative one equals Y minus negative. 5/2 equals zero minus six over negative three, Of course, zero minus six. Over negative three is equal to negative six over. Negative three or positive, too. And so we end up with two equations in terms of x and y So we have the X minus one is equal to negative two. And why plus five is equal to four. Solving the system of equations we have X is equal to negative one. And why equals negative one? And therefore our line intersects the X Y plane at the point. Negative one negative one zero and we'll do a similar procedure for the other planes. So intersects the X Z plane. This is when y equals zero. And so we obtained the equations X minus one over negative one equals zero minus negative. 5/2 equals Z minus six over negative three. And this equation in the middle reduces to five hats and so we obtain the system of equations. X minus one equals negative. Five halves and Z minus six equals negative 15 halves. Solving this is X equals negative. Three halves and Z equals. Let's see, also negative three halves. And so our line intersects the xz plane at negative three halves, zero negative. Three hats. All right. And finally you want to find where you line intersects the Y Z plane. Well, this happens when X is equal to zero. So plugging this into our symmetric equations we have zero minus one over negative one equals Y minus negative five over to equals two minus sorry Z minus six over negative three. And the expression on the left simplifies to negative one over negative one or just one. And so we have to change. The system of equations y plus five equals two and Z minus six equals negative three. Solving the system Why is equal to negative three and Z equals positive three. And so we obtain that the line intersects the plane at the point zero negative through three

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James Stewart

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Samuel Hannah

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Samuel Hannah

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Samuel Hannah

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals