a. Answer The set of parametric equations of the line passing through the given points is x = − 1 + 4 t , y = 3 + 3 t and z = 5 − 6 t Explanation Given: A line passing through the points ( − 1 , 3 , 5 ) and ( 3 , 6 , − 1 ) Concept Used: A line L passing through the point P = ( x 1 , y 1 , z 1 ) and parallel to the vector v = 〈 a , b , c 〉 is represented by the parametric equations x = x 1 + a t , y = y 1 + b t and z = z 1 + c t Calculation: Forthe line passing through the points ( − 1 , 3 , 5 ) and ( 3 , 6 , − 1 ) Let P = ( − 1 , 3 , 5 ) and Q = ( 3 , 6 , − 1 ) Then P = ( − 1 , 3 , 5 ) a direction vector for the line passing through P and Q is v = P Q → = 〈 3 + 1 , 6 − 3 , − 1 − 5 〉 = 〈 4 , 3 , − 6 〉 Using the coordinates of the initial point P = ( − 1 , 3 , 5 ) x 1 = − 1 , y 1 = 3 and z 1 = 5 And the direction numbers as a = 4 , b = 3 and c = − 6 The set of parametric equations of the line is x = − 1 + 4 t , y = 3 + 3 t and z = 5 − 6 t Conclusion: The set of parametric equations of the line passing through the given points is x = − 1 + 4 t , y = 3 + 3 t and z = 5 − 6 t b. To determine ToFind: A set of symmetric equations of the line passing through the points ( − 1 , 3 , 5 ) and ( 3 , 6 , − 1 ) Answer The set of symmetric equations of the line passing through the given points is x + 1 4 = y − 3 3 = z − 5 − 6 Explanation Given: A line passing through the points ( − 1 , 3 , 5 ) and ( 3 , 6 , − 1 ) Concept Used: A line L passing through the point P = ( x 1 , y 1 , z 1 ) and parallel to the vector v = 〈 a , b , c 〉 is represented by the parametric equations x = x 1 + a t , y = y 1 + b t and z = z 1 + c t When the direction numbers a , b and c are all nonzero, then symmetric equations of the line is given by x − x 1 a = y − y 1 b = z − z 1 c Calculation: For the line passing through the points ( − 1 , 3 , 5 ) and ( 3 , 6 , − 1 ) Let P = ( − 1 , 3 , 5 ) and Q = ( 3 , 6 , − 1 ) Then P = ( − 1 , 3 , 5 ) a direction vector for the line passing through P and Q is v = P Q → = 〈 3 + 1 , 6 − 3 , − 1 − 5 〉 = 〈 4 , 3 , − 6 〉 Using the coordinates of the initial point P = ( − 1 , 3 , 5 ) x 1 = − 1 , y 1 = 3 and z 1 = 5 And the direction numbers as a = 4 , b = 3 and c = − 6 Since, the direction numbers a , b and c are all nonzero, the symmetric equations of the line is given by x + 1 4 = y − 3 3 = z − 5 − 6 Conclusion: The set of symmetric equations of the line passing through the given points is x + 1 4 = y − 3 3 = z − 5 − 6

4th Edition

ISBN: 9781337516853

Author: Larson

Publisher: CENGAGE CO

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Question

Chapter 11, Problem 61RE

To determine

ToFind:A set of parametric equations of the line passing through the points $(−1,3,5)$ and $(3,6,−1)$

Expert Solution

Answer to Problem 61RE

The set of parametric equations of the line passing through the given points is

$x=−1+4t$ , $y=3+3t$ and $z=5−6t$

Explanation of Solution

Given:

A line passing through the points $(−1,3,5)$ and $(3,6,−1)$

Concept Used:

A line $L$ passing through the point $P=(x1,y1,z1)$ and parallel to the vector $v=〈a,b,c〉$ is represented by the parametric equations

$x=x1+at$ , $y=y1+bt$ and $z=z1+ct$

Calculation:

Forthe line passing through the points $(−1,3,5)$ and $(3,6,−1)$

Let $P=(−1,3,5)$ and $Q=(3,6,−1)$

Then $P=(−1,3,5)$ a direction vector for the line passing through $P$ and $Q$ is

$v=PQ→=〈3+1,6−3,−1−5〉=〈4,3,−6〉$

Using the coordinates of the initial point $P=(−1,3,5)$

$x1=−1$ , $y1=3$ and $z1=5$

And the direction numbers as

$a=4$ , $b=3$ and $c=−6$

The set of parametric equations of the line is

$x=−1+4t$ , $y=3+3t$ and $z=5−6t$

Conclusion:

The set of parametric equations of the line passing through the given points is

$x=−1+4t$ , $y=3+3t$ and $z=5−6t$

To determine

ToFind: A set of symmetric equations of the line passing through the points $(−1,3,5)$ and $(3,6,−1)$

Expert Solution

Answer to Problem 61RE

The set of symmetric equations of the line passing through the given points is

$x+14=y−33=z−5−6$

Explanation of Solution

Given:

A line passing through the points $(−1,3,5)$ and $(3,6,−1)$

Concept Used:

A line $L$ passing through the point $P=(x1,y1,z1)$ and parallel to the vector $v=〈a,b,c〉$ is represented by the parametric equations

$x=x1+at$ , $y=y1+bt$ and $z=z1+ct$

When the direction numbers $a$ , $b$ and $c$ are all nonzero, then symmetric equations of the line is given by

$x−x1a=y−y1b=z−z1c$

Calculation:

For the line passing through the points $(−1,3,5)$ and $(3,6,−1)$

Let $P=(−1,3,5)$ and $Q=(3,6,−1)$

Then $P=(−1,3,5)$ a direction vector for the line passing through $P$ and $Q$ is

$v=PQ→=〈3+1,6−3,−1−5〉=〈4,3,−6〉$

Using the coordinates of the initial point $P=(−1,3,5)$

$x1=−1$ , $y1=3$ and $z1=5$

And the direction numbers as

$a=4$ , $b=3$ and $c=−6$

Since, the direction numbers $a$ , $b$ and $c$ are all nonzero, the symmetric equations of the line is given by

$x+14=y−33=z−5−6$

Conclusion:

The set of symmetric equations of the line passing through the given points is

$x+14=y−33=z−5−6$

Chapter 11, Problem 60RE

Chapter 11, Problem 62RE

Chapter 11 Solutions

EBK PRECALCULUS W/LIMITS

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ToFind: A set of parametric equations of the line passing through the points ( − 1 , 3 , 5 ) and ( 3 , 6 , − 1 )

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Answer to Problem 61RE

Explanation of Solution

Answer to Problem 61RE

Explanation of Solution

Chapter 11 Solutions