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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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