EBK PRECALCULUS W/LIMITS

3rd Edition

ISBN: 9781285607160

Author: Larson

Publisher: CENGAGE CO

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Question

Chapter 11, Problem 17PS

To determine

Find the distance between the point and the line given by each set of parametric equations.

Expert Solution

Answer to Problem 17PS

$2.96units$

Explanation of Solution

Given information:

The distance between a point $Q$ and a line in space is given by

$D=‖PQ→×u‖‖u‖$

where $u$ is a direction vector for the line and $P$ is a point on the line. Find the distance between the point and the line given by each set of parametric equations.

$(1,5,−2)$

$x=−2+4t,y=3,z=1−t$

Calculation:

Consider the vectors in space $a=x1i+y1j+z1k$ and $b=x2i+y2j+z2k$ then the cross product of $a=x1i+y1j+z1k$ and $b=x2i+y2j+z2k$ will be the vector

$a×b=|ijkx1y1z1x2y2z2|$

$=i[y1(z2)−y2(z1)]−j[x1(z2)−x2(z1)]+k[x1(y2)−x2(y1)]$

Distance between a point Q and line in space with direction vector $u$ is

$D=‖PQ→×u‖‖u‖$

The coordinates of point Q is $(1,5,−2)$ . The parametric equation of line is

$x=−2+4ty=3z=1−t$

Direction vector of line is $u=〈4,0,−1〉$ . Now the coordinates of a general point on the line will be

$x=4λ−2y=3z=1−λ$

Hence coordinates of point $P$ is $(4λ−2,3,1−λ)$ .

$PQ→=〈4λ−2−1,3−5,1−λ+2〉=〈4λ−3,−2,3−λ〉$

Now cross product of two vectors and is

$PQ→×u=[ijk4λ−3−23−λ40−1]$

$=i(2−0)−j(−4λ+3−12+4λ)+(0+8)=i(2)−j(−9)+k(8)=〈2,9,8〉$

Therefore

$D=‖PQ→×u‖‖u‖$

$=22+92+8242+(−1)2=2.96$

Hence the distance of given point from the line is $2.96units$ .

To determine

Find the distance between the point and the line given by each set of parametric equations.

Expert Solution

Answer to Problem 17PS

$0.745units$

Explanation of Solution

Given information:

The distance between a point $Q$ and a line in space is given by

$D=‖PQ→×u‖‖u‖$

where $u$ is a direction vector for the line and $P$ is a point on the line. Find the distance between the point and the line given by each set of parametric equations.

$(1,−2,4)$

$x=2t,y=−3+t,z=2+2t$

Calculation:

Consider the vectors in space $a=x1i+y1j+z1k$ and $b=x2i+y2j+z2k$ then the cross product of $a=x1i+y1j+z1k$ and $b=x2i+y2j+z2k$ will be the vector

$a×b=|ijkx1y1z1x2y2z2|$

$=i[y1(z2)−y2(z1)]−j[x1(z2)−x2(z1)]+k[x1(y2)−x2(y1)]$

Distance between a point Q and line in space with direction vector $u$ is

$D=‖PQ→×u‖‖u‖$

The coordinates of point Q is $(1,−2,4)$ . The parametric equation of line is

$x=2ty=−3+tz=2+2t$

Direction vector of line is $u=〈2,1,2〉$ . Now the coordinates of a general point on the line will be

$x=2λy=−3+λz=2+2λ$

Therefore coordinates of point $P$ is $(2λ,−3+λ,2+2λ)$ . Hence vector $PQ→$ is

$PQ→=〈2λ−1,−3+λ+2,2+2λ−4〉=〈2λ−1,−1+λ,−2+2λ〉$

Now cross of two vectors and is

$PQ→×u=[ijk2λ−1−1+λ−2+2λ212]$

$=i(−2+2λ+2−2λ)−j(4λ−2+4−4λ)+k(2λ−1+2−2λ)=i(0)−j(−2)+k(1)=〈0,−2,1〉$

Therefore

$D=‖PQ→×u‖‖u‖$

$=(−2)2+1222+12+22=0.745$

Hence the distance of point $(1,−2,4)$ from the line $PQ$ is $0.745units$ .

Chapter 11, Problem 16PS

Chapter 11, Problem 18PS

Chapter 11 Solutions

EBK PRECALCULUS W/LIMITS

Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E

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