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# 1a)Find parametric equations of the line of the intersection of the pl.docx

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# 1a)Find parametric equations of the line of the intersection of the pl.docx

1a)Find parametric equations of the line of the intersection of the planes
x-2yplus 7z=0 and 3xplus 5y-z=0

1b) If lul and lvl are pependicular vectors such that lul=6 and lvl=4,
find the projection of 2u-3v on u plus 2v.
(I had to write the plus sign because it wasn\'t showing up when I posted my question.
Solution
as a example z = x + y 2x - 5y - z = 1 Let\'s recast the equations of the planes. x + y - z = 0 2x - 5y - z = 1 The cross product of the normal vectors of the two planes will be the directional vector v, of the line of intersection. v = <1, 1, -1> X <2, -5, -1> = <-6, -1, -7> Any non-zero multiple of v will also be a directional vector of the line. Multiply by -1. v = <6, 1, 7> Now find a point on the line. It will be a point in both planes. Let y = 0 and solve for x and z. x - z = 0 2x - z = 1 Subtract the first equation from the second. x = 1 Plug back into the first equation and solve for z. 1 - z = 0 z = 1 So our point on the line is P(1, 0, 1). The equation of the line of intersection is: L(t) = P + tv L(t) = <1, 0, 1> + t<6, 1, 7> where t is a scalar ranging over the real numbers Now put the equation of the line in parametric form. L(t): x = 1 + 6t y = t z = 1 + 7t __________ Both points (0, -1/6, -1/6) and (1/7, -1/7, 0) are on the line as is my point of P(1, 0, 1). In order for your parametric equation to be correct you need: 1) Any non-zero multiple of the directional vector of the line. 2) Any point on the line. So any of the three points, and an infinite number of other points on the line, along with a directional vector of the line will give you correct parametric equations. Your point is fine.
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1a)Find parametric equations of the line of the intersection of the planes
x-2yplus 7z=0 and 3xplus 5y-z=0

1b) If lul and lvl are pependicular vectors such that lul=6 and lvl=4,
find the projection of 2u-3v on u plus 2v.
(I had to write the plus sign because it wasn\'t showing up when I posted my question.
Solution
as a example z = x + y 2x - 5y - z = 1 Let\'s recast the equations of the planes. x + y - z = 0 2x - 5y - z = 1 The cross product of the normal vectors of the two planes will be the directional vector v, of the line of intersection. v = <1, 1, -1> X <2, -5, -1> = <-6, -1, -7> Any non-zero multiple of v will also be a directional vector of the line. Multiply by -1. v = <6, 1, 7> Now find a point on the line. It will be a point in both planes. Let y = 0 and solve for x and z. x - z = 0 2x - z = 1 Subtract the first equation from the second. x = 1 Plug back into the first equation and solve for z. 1 - z = 0 z = 1 So our point on the line is P(1, 0, 1). The equation of the line of intersection is: L(t) = P + tv L(t) = <1, 0, 1> + t<6, 1, 7> where t is a scalar ranging over the real numbers Now put the equation of the line in parametric form. L(t): x = 1 + 6t y = t z = 1 + 7t __________ Both points (0, -1/6, -1/6) and (1/7, -1/7, 0) are on the line as is my point of P(1, 0, 1). In order for your parametric equation to be correct you need: 1) Any non-zero multiple of the directional vector of the line. 2) Any point on the line. So any of the three points, and an infinite number of other points on the line, along with a directional vector of the line will give you correct parametric equations. Your point is fine.
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### 1a)Find parametric equations of the line of the intersection of the pl.docx

1. 1. 1a)Find parametric equations of the line of the intersection of the planes x-2yplus 7z=0 and 3xplus 5y-z=0 1b) If lul and lvl are pependicular vectors such that lul=6 and lvl=4, find the projection of 2u-3v on u plus 2v. (I had to write the plus sign because it wasn't showing up when I posted my question. Solution as a example z = x + y 2x - 5y - z = 1 Let's recast the equations of the planes. x + y - z = 0 2x - 5y - z = 1 The cross product of the normal vectors of the two planes will be the directional vector v, of the line of intersection. v = <1, 1, -1> X <2, -5, -1> = <-6, -1, -7> Any non-zero multiple of v will also be a directional vector of the line. Multiply by -1. v = <6, 1, 7> Now find a point on the line. It will be a point in both planes. Let y = 0 and solve for x and z. x - z = 0 2x - z = 1 Subtract the first equation from the second. x = 1 Plug back into the first equation and solve for z. 1 - z = 0 z = 1 So our point on the line is P(1, 0, 1). The equation of the line of intersection is: L(t) = P + tv L(t) = <1, 0, 1> + t<6, 1, 7> where t is a scalar ranging over the real numbers Now put the equation of the line in parametric form. L(t): x = 1 + 6t y = t z = 1 + 7t __________ Both points (0, -1/6, -1/6) and (1/7, -1/7, 0) are on the line as is my point of P(1, 0, 1). In order for your parametric equation to be correct you need: 1) Any non-zero multiple of the directional vector of the line. 2) Any point on the line. So any of the three points, and an infinite number of other points on the line, along with a directional vector of the line will give you correct parametric equations. Your point is fine.