Find the distance of the point 2-3-5 from the plane r- -2 -2 k-9-

Equation of plane is x + 2y – 2z – 9 = 0 , s = |ax_1 + by_1 + cz_1 + d|/√(a^2 + b^2 + c^2) = |2+6 10 9|/√(1+4+4)=11/3units ...

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Distance Formula

Find the dist...

Question

Find the distance of the point (2, 3, -5) from the plane $\to r . (^i + 2^j - 2^k) = 9$ .

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Solution

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\to r (^i + 2^j +^k) = 4

\to r (^- i +^j + 2^k) = 9

\to r = 6^i + 2^j + 2^k + λ (^i - 2^j + 2^k)

\to r = - 4^i -^k + μ (3^i - 2^j - 2^k)

\to r =^i + 2^j +^k + λ (^i -^j +^k)

\to r = 2^i -^j -^k + μ (2^i +^j + 2^k)

\to r = (4^i -^j) + λ (^i + 2^j - 3^k)

\to r = (^i -^j + 2^k) + μ (^i + 4^j - 5^k)

(2, 12, 5)

\to r = 2^i - 4^j + 2^k + λ (3^i + 4^j + 2^k)

\to r (^i - 2^j +^k) = 0

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