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Question
Solve the following :
Find the vector equation of the plane which makes equal non zero intercepts on the coordinate axes and passes through (1, 1, 1).
Solution
Case 1 : Let all the intercepts be 0.
Then the plane passes through the origin.
Then the vector equation of the plane is ax + by + cz = 0. ...(1)
(1, 1, 1) lie on the plane.
∴ 1a + 1b + 1c = 0
∴ `hat"i"/|(1, 1),(1, 1)| = hat"j"/|(1, 1),(1, 1)| = hat"k"/|(1, 1),(1, 1)|`
∴ `hat"i"/(1) = hat"j"/(1) = hat"k"/(1)`
i.e. `hat"i"/(1) = hat"j"/(1) = hat"k"/(1)`
∴ `hat"i", hat"j", hat"k"` are proprtional to 1, 1, 1
∴ from (1), the required cartesian equation is x - y + z = 0
Case 2 : Let he plane make non zero intercept p on each axis.
then its equation is `hat"i"/p + hat"j"/p + hat"k"/p` = 1
i.e. `hat"i" + hat"j" + hat"k"` = p ...(2)
Since this plane pass through (1, 1, 1)
∴ 1 + 1 + 1 = p
∴ p = 3
∴ from (2), the required cartesian equation is `hat"i" + hat"j" + hat"k"` = 3
Hence, the cartesian equations of required planes are
`bar"r".(hat"i" + hat"j" + hat"k")` = 3
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