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Question
Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.
Solution
It is given that,
Intercepts are a = 3, b = −4, c = 2
The intercept form of a plane is as follows:
`x/a+y/b+z/c=1`
`therefore x/3+y/-4+z/2=1`
The equation of the given plane is
4x − 3y + 6z = 12.
`(xhati+yhatj+zhatk).(4hati-3hatj+6hatk)=12`
`vec r.(4 hati-3hatj+6hatk)=12`
This is the vector form of the equation of the given plane.
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