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Dweath50
16.01.2020 •
Mathematics
Find the equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, – 1, 5).
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Ответ:
x+5y-6z+18 =0
Step-by-step explanation:
We have to find the equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points A(3, 4, – 1) and B(2, – 1, 5).
The perpendicular line joining AB is normal to the plane.
Direction ratios of AB are =![(2-3, -1-4, 5-(-1))\\=(-1, -5, 6)](/tpl/images/0457/9112/d37ca.png)
this is direction ratios of normal
The plane passes through the point (3,-3,1)
Hence equation of the plane in normal form is
x+5y-6z+18 =0 is the equation of the plane.
Ответ: