Question

Find the equation of the plane through the points (2 -1 0) and (3 -4 5) and parallel to the line 2x=3y= 4z vector and Cartesian form.​

Answer 1

Step-by-step explanation:

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Answer 2

The equation of the plane passing through the point A(2,-1,0) and B(3,-4,5) and paallel to 2x=3y=4z is 29(x-2)-27(y+1)+14(z-0)=0

1.Direction ratios of 2x=3y=4z are 6,4,3.

2.Direction ratios of the line segment AB is (3-2),(-4+1),(5-0) are 1,-3,5

3.these two direction ratios are perpendicular so

6a+4b+3c=0.......1

And a-3b+5c=0.........2

4.from 1 and 2 we get a/20+9=b/-30+3=c/18-4

=a/29=b/-27=c/14

Therefore the direction ratios of plane are 29,-27,14

And equation of plane considering point A (2,-1,0) is

29(x-2)-27(y+1)+14(z-0)=0