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 November 22nd, 2011, 05:55 PM #1 Member   Joined: Sep 2011 Posts: 49 Thanks: 0 Find the equation of the line through P that cuts the line? Find the equation of the line through P(-1,0,1) that cuts the line r=(3,2,1)+t(1,2,2) at right-angles at Q Also find the length PQ and the equation of the plane containing the two lines. Given that a=(3,2,1) and b=(1,-2,-4) are the position vectors of the points P and Q respectively, find a)the equation of the plane passing through Q and the perpendicular to PQ b)the distance from the point (-1,1,1)to the plane in (a)
November 23rd, 2011, 03:19 PM   #2
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Re: Find the equation of the line through P that cuts the li

Hello, luketapis!

Here's the first one . . .

Quote:
 $\text{(a) Find the equation of the line through }P(-1,\,0,\,1)$ [color=beige]. . . [/color]$\text{that cuts the line }r\:=\3,\,2,\,1)\,+\,t(1,\,2,\,2)\text{ at right angles at }Q." />

$\text{W\!e have a point }P(-1,\,0,\,1) \;\text{ and the line }L_1:\;\begin{Bmatrix}x=&3+t \\ y=&2+2t \\ z=&1+2t \end{Bmatrix}=$

$\text{Point }Q\text{ is on line }L_1\text{, where: }\,Q(3+t,\,2+2t,\,1+2t)$

$PQ\text{ is perpendicular to }L_1\text{ when the distance }\overline{PQ}\text{ is a minimum.}$

[color=beige]. . [/color]$\overline{PQ}^2 \;=\;\big([3+t]\,-\,[-1]\big)^2\,+\,\big([2+2t]\,-\,0\big)^2 + \big([1+2t]\,-\,1\big)^2$

[color=beige]. . [/color]$\overline{PQ}^2 \;=\;9t^2\,+\,16t\,+\,20\;\;*[1]*$

$\text{The function is an up-opening parabola.}$
[color=beige]. . [/color]$\text{Its minimum value is at its }vertex:\;v \,=\,\frac{-b}{2a}$
[color=beige]. . . . [/color]$t \:=\:\frac{-16}{2(9)} \:=\:-\frac{8}{9}$

$\text{Hence, }Q\text{ is: }\;\begin{Bmatrix}x=&3-\frac{8}{9}=&\frac{19}{9} \\ \\ \\ y=&2-\frac{16}{9}=&\frac{2}{9} \\ \\ \\ z=&1 - \frac{16}{9}=&-\frac{7}{9} \end{Bmatrix} \;\;\;\Rightarrow\;\;\;Q\left(\frac{19}{9},\,\frac {2}{9},\,-\frac{7}{9}\right)=$

$\text{The vector }\vec v \:=\:\vec{PQ} \;=\;\left\langle\frac{19}{9}\,-\,(-1),\:\frac{2}{9}\,-\,0,\:-\frac{7}{9}\,-\,1\right\rangle \;=\;\left\langle \frac{28}{9},\:\frac{2}{9},\:-\frac{16}{9} \right\rangle$

[color=beige]. . [/color]$\text{Hence: }\:\vec v \:=\:\langle 14,\:1,\:-8\rangle$

$\text{The line through }P(-1,\,0,\,1)\text{ with direction }\vec v \:=\:\langle 14,\,1,-8\rangle\text{ is:}$

[color=beige]. . [/color]$L_2:\;\begin{Bmatrix}x=&-1\,+\,14u \\ \\ \\ u=&0\,+\,u \\ \\ \\ z=&1\,-\,8u\end{Bmatrix}=$

Quote:
 (b) Find the length $PQ.$

$\text{Substitute }t= -\frac{8}{9}\text{ into }*[1]*:$

$\overline{PQ}^2 \:=\:9\left(-\frac{8}{9}\right)^2\,+\,16\left(-\frac{8}{9}\right)\,+\,20 \;=\;\frac{116}{9}$

$\text{Therefore: }\:\overline{PQ} \:=\:\sqrt{\frac{116}{9}} \;=\;\frac{2}{3}\sqrt{29}$

Quote:
 (c) Find the equation of the plane containing the two lines.

$\text{The lines are: }\;L_1:\:\begin{Bmatrix}x=&3\,+\,t \\ \\ y=&2\,+\,2t \\ \\ z=&1\,+\,2t \end{Bmatrix} \;\text{ and }\;L_2:\;\begin{Bmatrix}x=&-1\,+\,14u \\ \\ y=&0\,+\,u \\ \\ z=&1\,-\,8u \end{Bmatrix}=$

$\text{W\!e have direction vectors: }\:\vec v_1= \langle1,\,2,\,2\rangle\,\text{ and }\,\vec v_2 = \langle 14,\,1,\,8\rangle$

$\text{The normal direction of the plane is:}$
[color=beige]. . [/color]$\vec n \;=\;\vec v_1\,\times\,\vec v_2 \;=\;\begin{vmatrix}i &j=&k \\ 1=&2=&2 \\ \\ \\ 14=&1=&-8 \end{vmatrix} \;=\; -18i\,+\,36j\,-\,27k \;=\;\langle 2.\,-4,\,3\rangle$

$\text{From part (a), point }Q\left(\frac{19}{9},\:\frac{2}{9},\:-\frac{7}{9}\right)\text{ is on the plane.}$

$\text{The plane through }\left(\frac{19}{9},\:\frac{2}{9},\:-\frac{7}{9}\right)\text{ with normal }\vec n \:=\:\langle 2,\,-4,\,3\rangle\text{ is:}$

[color=beige]. . [/color]$2\left(x\,-\,\frac{19}{9}\right)\,-\,4\left(y\,-\,\frac{2}{9}\right)\,+\,3\left(z\,+\,\frac{7}{9}\ right) \:=\:0$

[color=beige]. . [/color]$2x\,-\,\frac{38}{9}\,-\,4y\,+\,\frac{8}{9}\,+\,3z\,+\,\frac{21}{9} \:=\:0$

[color=beige]. . [/color]$2x\,-\,4y\,+\,3z\,-\,\frac{9}{9} \:=\:0$

[color=beige]. . [/color]$2x\,-\,4y\,+\,3z \;=\;1$

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