The vector equation of the plane which is at a distance of 3-38 from the origin and the normal from the origin is 5 -2 -3 k- is A- r -5 -2 -3 k-9B- r -5 -2 -3 k-3C- r -5 -2 -3 k-9-38D- r-5 -2 -3 k-9

The correct option is B r nbsp;· nbsp;(5 î 2 ĵ+3 k̂)=3 nbsp; nbsp;n=5 î 2 ĵ+3 k̂ ∴n̂=n|n|=5 î 2 ĵ+3 k̂√(38) Hence, the required of the plane is nbsp ...

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Standard XII

Mathematics

Equation of a Plane : Normal Form

The vector eq...

Question

The vector equation of the plane which is at a distance of $\frac{3}{\sqrt{38}}$ from the origin and the normal from the origin is $5^i - 2^j + 3^k$ is

\to r \cdot (5^i - 2^j + 3^k) = 9

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\to r \cdot (5^i - 2^j + 3^k) = 3

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\to r \cdot (5^i - 2^j + 3^k) = \frac{9}{38}

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\to r \cdot (- 5^i + 2^j - 3^k) = 9

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Solution

The correct option is B $\to r \cdot (5^i - 2^j + 3^k) = 3$
$\to n = 5^i - 2^j + 3^k$
$∴^n = \frac{\to n}{| \to n |} = \frac{5^i - 2^j + 3^k}{\sqrt{38}}$
Hence, the required of the plane is
$\to r \cdot (\frac{5^i - 2^j + 3^k}{\sqrt{38}}) = \frac{3}{\sqrt{38}}$
$\Rightarrow \to r \cdot (5^i - 2^j + 3^k) = 3$

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The vector equation of the plane which is at a distance of 3√38 from the origin and the normal from the origin is 5^i−2^j+3^k is

The correct option is B →r⋅(5^i−2^j+3^k)=3 →n=5^i−2^j+3^k ∴^n=→n|→n|=5^i−2^j+3^k√38 Hence, the required of the plane is →r⋅(5^i−2^j+3^k√38)=3√38 ⇒→r⋅(5^i−2^j+3^k)=3

The vector equation of the plane which is at a distance of $\frac{3}{\sqrt{38}}$ from the origin and the normal from the origin is $5^i - 2^j + 3^k$ is