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Marina86 [1]
3 years ago
7

Find a vector equation and parametric equations for the line through the point (1,0,6) and perpendicular to the plane x+3y+z=5.

Mathematics
1 answer:
tester [92]3 years ago
5 0

The normal vector to the plane <em>x</em> + 3<em>y</em> + <em>z</em> = 5 is <em>n</em> = (1, 3, 1). The line we want is parallel to this normal vector.

Scale this normal vector by any real number <em>t</em> to get the equation of the line through the point (1, 3, 1) and the origin, then translate it by the vector (1, 0, 6) to get the equation of the line we want:

(1, 0, 6) + (1, 3, 1)<em>t</em> = (1 + <em>t</em>, 3<em>t</em>, 6 + <em>t</em>)

This is the vector equation; getting the parametric form is just a matter of delineating

<em>x</em>(<em>t</em>) = 1 + <em>t</em>

<em>y</em>(<em>t</em>) = 3<em>t</em>

<em>z</em>(<em>t</em>) = 6 + <em>t</em>

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(Problem on image.)...
ANEK [815]
Hey there! :) 

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That equation is : a² + b² = c²

a = adjacent side

b = long side of rectangle

c = hypotenuse

We're already given the values of both a & c, so let's plug everything in! 

a² + b² = c²  → a = 10 , b = ? , c = 20

(10²) + b² = (20²)

Simplify.

100 + b² = 400

Subtract 100 from both sides.

b² = 400 - 100

Simplify.

b² = 300

Get the square root of b² & 300.

\sqrt{b^2} =  \sqrt{300}

Simplify!

b = 17.3

So, the missing side length is the first answer choice : 17.3

~Hope I helped!~
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Answer:

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7 0
4 years ago
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4 0
4 years ago
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