Find the angle between the lines

[zorro]

Homework Statement

I have 2 doubts-

1)The equation of the plane containing the line 2x+z-4=0=2y+z and passing through the point (2,1,-1) is?

2)Find the angle between the lines
(x-2)/3 = (y+1)/-2, z=2 and
(x-1)/1 = (2y+3)/3 = (z+5)/2

The Attempt at a Solution

1)How do I interpret the given equation of the line?
How to reduce it to symmetric form?
Is it 2x-4 = 2y = 0? (z cancels out)

2) In the first equation, there is a 'comma' between (x-2)/3 = (y+1)/-2 and z=2. What does it mean?
Moreover, z=2 is a plane not a line.
So do we have to find the angle between (x-2)/3 = (y+1)/-2 = z/1 and (x-1)/1 = (2y+3)/3 = (z+5)/2?

 

[madmike159]

To find the angle between 2 vectors you use the dot product.

 

[HallsofIvy]

Homework Statement

I have 2 doubts-

1)The equation of the plane containing the line 2x+z-4=0=2y+z and passing through the point (2,1,-1) is?

2)Find the angle between the lines
(x-2)/3 = (y+1)/-2, z=2 and
(x-1)/1 = (2y+3)/3 = (z+5)/2

The Attempt at a Solution

1)How do I interpret the given equation of the line?
How to reduce it to symmetric form?
Is it 2x-4 = 2y = 0? (z cancels out)

You cannot just cancel z in two of the equations. "Canceling z" here really means "subtract z from both sides" but since there are three "sides" that would give
2x- 4= 2y= -z. You really only need two points on that line: if You take z= 0, then 2y= 0 gives y= 0 and 2x- 4= 0 gives x= 2. (2, 0, 0) is a point on the line. If you take z= 2, then 2y= -2 gives y= -1 and 2x- 4= -2 gives 2x= 2 or x= 1. (1, -1, 2) is another point on the line.
Can you find the plane containing the three points (2, 0, 0), (1, -1, 2), and (2, 1, -1)?

2) In the first equation, there is a 'comma' between (x-2)/3 = (y+1)/-2 and z=2. What does it mean?
Moreover, z=2 is a plane not a line.

No, but the set of all points, (x, y, z), satisfying (x- 2)/3= (y+ 1)/(-2) and z= 2 is a line. Setting "t" equal to the common value of (x- 2)/3 and (y+ 1)/(-2), we have x= 3t+ 2 and y= -2t- 1. z= 2= 2+ 0t for all t. That gives as a vector pointing in the direction of the line $\vec{u}= 3\vec{i}- 2\vec{j}+ 0\vec{k}$.

If you set each part of (x-1)/1 = (2y+3)/3 = (z+5)/2 equal to the parameter s, you get x= s+ 1, 2y= 3s- 3 so y= (3/2)s- 1, and z= 2s+ 5. That has direction vector $\vec{v}= \vec{i}+ (3/2)\vec{j}+ 2\vec{k}$.

In general, if the "symmetric" equations are $\frac{x- a}{A}= \frac{y- b}{B}= \frac{z- c}{C}$, then a direction vector is $A\vec{i}+ B\vec{j}+ C\vec{k}$. If one of x, y, or z is given as a constant, that is the same as the denominator being 0.

So do we have to find the angle between (x-2)/3 = (y+1)/-2 = z/1

No, there is no "z/1", it would be better to think of it as "z/0" which, of course, does not exist- that's why z was not written that way.

and (x-1)/1 = (2y+3)/3 = (z+5)/2?

Of course, after you have the direction vectors, $\vec{u}$ and $\vec{v}$, you can find the angle between them from $\vec{u}\cdot\vec{v}= ||\vec{u}||||\vec{v}|| cos(\theta)$.

 

[zorro]

thanks!