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Find an equation of the plane.

The plane that passes through the points $ (0, -2, 5) $ and $ (-1, 3, 1) $ and is perpendicular to the plane $ 2z = 5x + 4y $

$-6 x+22 y+29 z=101$

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in the question they're asking to find the creation of the plane that passes through two points. A 0 -2, 5 and B -131. And This plane is also perpendicular to another plane. Having equation to share equal to five. Express for way. So we can rearrange this equation as five x plus four by -2. That is equal to zero since this plane is perpendicular to the plane asking the questions. So it's right. Yeah Victor is also perpendicular to the plane asking the question. So it's vector is equal to B. two vectors suppose is equal to by Icap. Let's for Jacob minus to kick up and in order to find b other victor from the points that is present in the plane given by a N. B. We can find another vector to the plane that is a B vector is equal to mhm -1 0 Icap Plus three plus 2 Jacob Plus 1 -5 K Cup. So this is equal to -1 icap plus five Jacob -4 K Cups. Yeah. Therefore mhm victor. We won is equal to every victor. Let's suppose so. Since this is the perpendicular to the plane. So A.B. or be one Victor is parallel to be to victor. Therefore the cross productive be one NV two will give the normal vector to the plane. Yeah that is equal to Icap Jacob kick up and the values are minus one five minus four, five, four minus two. This is equal to mhm six. Icap -20 to Jacob -29 K Cup. In order to find the equation of the required plane asking the question Yeah. Equation of the plane Can be written as six. Yeah. Okay X minus. Let us consider the point that was given initially as Yeah. Okay course zero minus 25. And we put this value in the place of finding the equation of the plane so it is x minus zero -22. y plus two minus 29. 0 -5 is equal to zero. And after solving this equation the equation becomes minus six X Last 22 way Plus 29 Z Is equal to 101. Okay and so this is the required answer to the given question which is the equation on the plane.