Find an equation of the plane. the plane through the point (−5, 9, 10) and perpendicular to the line x = 1 + t, y = 4t, z = 2 − 3t

The direction vector of the line
L: x=1+t, y=4t, z=2-3t
is <1,4,-3>
which is also the required normal vector of the plane.

Since the plane passes through point (-5,9,10), the required plane is :
Π 1(x-(-5)+4(y-9)-3(z-10)=0
=>
Π x+4y-3z=1

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topeadeniran2 topeadeniran2 · Answer 2 · 2021-09-09T15:52:24+02:00

The equation of the plane will be [tex]x + 4y - 3z = 1[/tex]

First and foremost, since we are given the information that the plane passes through the point [tex](-5, 9, 10)[/tex] and is then perpendicular to the line x = [tex]1 + t, y = 4t, z = 2 - 3t.[/tex]

Based on the above information, the direction vector of the line will be 1, 4, -3 which are the coefficients of t.

Since the plane goes through point [tex](-5,9,10)[/tex], therefore the required plane will be:

[tex]1[x -(-5)] + 4(y - 9) - 3(z - 10) = 0[/tex]

Then, we'll solve this equation which will be:

[tex]1(x + 5) + 4(y - 9) - 3(z - 10) = 0[/tex]

[tex]x + 5 + 4y - 36 - 3z + 30 = 0[/tex]

Collect like terms

[tex]x + 4y - 3z = 0 - 5 + 36 - 30.[/tex]

[tex]x + 4y - 3z = 1[/tex]

Therefore, the equation of the plane will be [tex]x + 4y - 3z = 1[/tex]

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