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>
Answer:
By the angles and sides, but if you need more help here is a link to a video that could be pretty helpful.
Step-by-step explanation:
https://www.khanacademy.org/math/cc-fourth-grade-math/plane-figures/imp-classifying-triangles/v/scalene-isosceles-equilateral-acute-right-obtuse#:~:text=Learn%20to%20categorize%20triangles%20as,acute%2C%20right%2C%20or%20obtuse.
Answer:
C= 2.18x + 2.39
Step-by-step explanation:
Try the case if the triangle was a right triangle.
Then the longest side, 25, would be c or the hypotenuse.
9^2+24^2=25^2
81+576=625
657=625
This is not a true statement so the triangle is not a right triangle.
Hope this helps :)
