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il63 [147K]
3 years ago
6

Find a set of scalar parametric equations for the line formed by the two intersecting planes. 3x+3y+2z+2=0 2x−3y+2z−2=0

Mathematics
1 answer:
vivado [14]3 years ago
8 0

The normal vectors to the two planes are (3, 3, 2) and (2, -3, 2). The cross product of these will be the direction vector of the line of intersection, (12, -2, -15).

Using x=0, we can find a point on this line by solving the simultaneous equations that remain:

... 3y +2z = -2

... -3y +2z = 2

Adding these, we get

... 4z = 0

... z = 0

so the point we're looking for is (x, y, z) = (0, -2/3, 0). This gives rise to the parametric equations ...

  • x = 12t
  • y = -2/3 -2t
  • z = -15t

By letting t=2/3, we can find a point on the line that has integer coefficients. That will be (x, y, z) = (8, -2, -10).

Then our parametric equations can be written as

  • x = 8 +12t
  • y = -2 -2t
  • z = -10 -15t
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Answer:

(1\frac{1}{2},1) - False

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Step-by-step explanation:

The points are

(1\frac{1}{2},1) , (4,6), (18,12), (0,0) and (2\frac{1}{2},3\frac{3}{4}) ---- missing from the question

Given

y = 1\frac{1}{2}x

Required

Determine if each of the points would be on y = 1\frac{1}{2}x

To do this, we simply substitute the value of x and of each point in y = 1\frac{1}{2}x.

(a) (1\frac{1}{2},1)

In this case;

x = 1\frac{1}{2} and y = 1

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 1\frac{1}{2}

y = \frac{3}{2} * \frac{3}{2}

y = \frac{9}{4}

y = 2\frac{1}{4}

<em>The point </em>(1\frac{1}{2},1)<em>  won't be on the graph because the corresponding value of y for </em>x = 1\frac{1}{2}<em> is </em>y = 2\frac{1}{4}<em></em>

(b) (4,6)

In this case;

x = 4

y = 6

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 4

y = \frac{3}{2} * 4

y = \frac{3* 4}{2}

y = \frac{12}{2}

y = 6

<em>The point </em>(4,6)<em>  would be on the graph because the corresponding value of y for </em>x = 4 is y = 6

(c) (18,12)

In this case:

x = 18;y = 12

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 18

y = \frac{3}{2} * 18

y = \frac{3* 18}{2}

y = \frac{54}{2}

y = 27

<em>The point </em>(18,12)<em>  wouldn't be on the graph because the corresponding value of y for </em>x = 18<em> is </em>y = 12<em></em>

(d) (0,0)

In this case;

x =0; y = 0

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 0

y = 0

<em>The point </em>(0,0)<em>  would be on the graph because the corresponding value of y for </em>x = 0 is y = 0

(e) (2\frac{1}{2},3\frac{3}{4})

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x = 2\frac{1}{2}; y = 3\frac{3}{4}

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y = 1\frac{1}{2} * 2\frac{1}{2}

y = \frac{3}{2} * \frac{5}{2}

y = \frac{15}{4}

y = 3\frac{3}{4}

<em>The point </em>(2\frac{1}{2},3\frac{3}{4}) <em>  would be on the graph because the corresponding value of y for </em>x = 2\frac{1}{2} is y = 3\frac{3}{4}

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