Equations of Lines in 3D

Let l be the line represented by the symmetric equations

\displaystyle\frac{x-2}{4} = \frac{y+3}{-2}=\frac{z-1}{7}

1. Find the point of intersection of l and the plane 5x-y+2z=12.

2. Find the distance between P(0,1,0) and l.

3. The acute angle between l and the line m: \hspace{.3in} x=1+t; \hspace{.2in} y=3-2t; \hspace{.2in} z=4t.

Solution:

1. Note that transforming l in its parametric form of equations gives us:

x=4t+2; \hspace{.2in} y=2t-3; \hspace{.2in} z=7t+1

Plugging the values x, y and z into the equation of the plane gives us an equation of one variable t:

5(4t+2)-(-2t-3)+2(7t+1)=12

Solving the above equation gives us the value t=-1/12.

Finally, plug this value of parameter to the equation of the line.

x=4(-1/12)+2 = 5/3

y=-2(-1/12)-3 = -17/6

z=7(-1/12)+1 = 5/12

Thus, the point of intersection is the point (5/3,-17/6,5/12).

2. A parallel vector to l is <4,-2,7> =: R and a point on l is Q(2,-3,1). (It is hard to put a drawing here so, just do the visualization yourself). The plan is to construct a right triangle with hypotenuse parallel to the vector PQ and one leg parallel to the R . We will find now the scalar projection of PQ onto R and the magnitude of PQ.

sp(PQ,R) = \displaystyle\frac{<2,-4,1> \cdot <4,-2,7>}{ \sqrt{16+4+49}} = \frac{23}{\sqrt{69}}

\left\| PQ \right\| = \sqrt{4+16+1}= \sqrt{21}

This is still the length of the leg parallel to R and the length of the hypotenuse. FInally, we use Pythagorean Theorem to find our desired distance.

dist(P,l) = \sqrt{21 - \frac{23^2}{69}}=\sqrt{\frac{920}{69}}

3. The parallel vector to l is <4,-2,7> =: R while the parallel vector to m is <1,-2,4> =: S . The acute angle is then found from the formula of the dot product

\theta = \arccos\displaystyle\frac{R\cdot S}{\left\|R\right\| \cdot \left\| S \right\|}

Meanwhile,

R\cdot S = <4,-2,7> \cdot <1,-2,4> = 36

\left\|R\right\| = \sqrt{16+4+49}=\sqrt{69}

\left\| S \right\| = \sqrt{1+4+16} = \sqrt{21}

Plugging all of these values gives us

\theta = \arccos\displaystyle\frac {36}{\sqrt{69} \cdot \sqrt{21}}

This entry was posted on Wednesday, April 30th, 2008 at 6:38 pm and is filed under Calculus, Geometry. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “Equations of Lines in 3D”

  1. Sri Yuliani Says:
    December 10, 2008 at 4:03 am | Reply

    Thank, this will be useful for my daughters in solving their homework

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