The vector equation of the plane which is at a distance of 3-14 from the origin and the normal from the origin is 2 -3 -k- isA- r-2 -3 -k-3B- r -2 -3C- r -k-9D- r -2 -k-3

The correct option is A r·(2î 3ĵ+k̂)=3d=3/√(14) n=2î 3ĵ+k̂ ∴n̂=2î 3ĵ+k̂/√(14) Hence, the required equation of the plane is nbsp; r·n̂=d ⇒ nbsp;r· nbsp;(2 ...

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Byju's Answer

Standard XII

Physics

Unit Vectors

The vector eq...

Question

The vector equation of the plane which is at a distance of $\frac{3}{\sqrt{14}}$ from the origin and the normal from the origin is $2^i - 3^j +^k$ is

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

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

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

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

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Solution

The correct option is A $\to r \cdot (2^i - 3^j +^k) = 3$
$d = \frac{3}{\sqrt{14}}$
$\to n = 2^i - 3^j +^k$
$∴^n = \frac{2^i - 3^j +^k}{\sqrt{14}}$
Hence, the required equation of the plane is $\to r \cdot^n = d$
$\Rightarrow \to r \cdot (2^i - 3^j +^k) = 3$

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Similar questions

Q. The unit vector in the direction of

2^i + 3^j +^k

Q. The vector equation of the plane which is at a distance of

\frac{3}{\sqrt{14}}

from the origin and the normal from the origin is

2^i - 3^j +^k

Q. Find the vector

\to r

which is perpendicular to

\to a =^i - 2^j + 5^k

and

\to b = 2^i + 3^j -^k

and

\to r . (2^i +^j +^k) + 8 = 0

Q. The distance between the plane

\to r . (^i + 5^j +^k) = 5

and the line

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

Q. A unit vector perpendicular to the plane containing the vector

^i + 2^j +^k

and

- 2^i +^j + 3^k

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The vector equation of the plane which is at a distance of 3√14 from the origin and the normal from the origin is 2^i−3^j+^k is

The correct option is A →r⋅(2^i−3^j+^k)=3d=3√14 →n=2^i−3^j+^k ∴^n=2^i−3^j+^k√14 Hence, the required equation of the plane is →r⋅^n=d ⇒→r⋅(2^i−3^j+^k)=3

The vector equation of the plane which is at a distance of $\frac{3}{\sqrt{14}}$ from the origin and the normal from the origin is $2^i - 3^j +^k$ is

The correct option is A $\to r \cdot (2^i - 3^j +^k) = 3$
$d = \frac{3}{\sqrt{14}}$
$\to n = 2^i - 3^j +^k$
$∴^n = \frac{2^i - 3^j +^k}{\sqrt{14}}$
Hence, the required equation of the plane is $\to r \cdot^n = d$
$\Rightarrow \to r \cdot (2^i - 3^j +^k) = 3$