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Problem 25 Easy Difficulty

Find an equation of the plane.

The plane through the point $ (-1, \frac{1}{2}, 3) $ and with normal vector $ i + 4j + k $

Answer

$x+4 y+z=4$

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Video Transcript

Hello. So the question is taken home vectors and geometry of the space where we hear we I need to find the equation of plain that plane passing two minus 11 by two and three. And normal vector I plus four plus case. So given the point is minus one. One by two and three. And it's normal victories. Mhm. I plus four G plus K. Okay so we know that uh equation of plane. Sure. Yeah passing to what? Passing two X zero Y zero and there's zero. And uh it is normal to victor Ai plus B. J plus see case A into -X0. Let's be into Y -Y0 plus C. Into k minus K. Zero is equal to zero using the same form here. So we get What explains went into one plus Fall into Y -1 x two Plus She's three in 2. So he went into 30 -3 that will visit. So if you're dead Zed minus that zero that is equal to zero. The value of C. Is one minus that. Zero is three. That is equal to zero. So that equation will be X plus one plus four. Y minus two plus zero minus three is equal to zero. So from here the value of equation is expressed for white list said is equal to minus five plus one is Plus 4. 1. That I can say which is the required the creation of plain that passing to that point and it's normal to their director. So hope this close your doubt and thank