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Question
O is the origin and A is (a, b, c). Find the direction cosines of the line OA and the equation of plane through A at right angle to OA.
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Solution
Given, O(0, 0, 0) and A(a, b, c)
So, the direction ratios of OA = a – 0, b – 0, c – 0 = a, b, c
And, the direction cosines of line OA
`(a/(sqrt(a^2 + b^2 + c^2)), b/sqrt(a^2 + b^2 + c^2), c/sqrt(a^2 + b^2 + c^2))`
Now, the direction ratios of the normal to the plane are (a, b, c).
We know that, the equation of the plan passing through the point A(a, b, c) is
a(x – a) + b(y – b) + c(z – c) = 0
ax – a2 + by – b2 + cz – c2 = 0
ax + by + cz = a2 + b2 + c2
Thus, the required equation of the plane is ax + by + cz = a2 + b2 + c2
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