OABC is a tetrahedron- The position vectors of A- B and C are - - -k- respectively- -O- is the origin- The distance of the plane face ABC from origin O is

The correct option is A 1/√(2) Volume of tetrahedron =1/6[ OA OB OC ]=1/61 nbsp; amp; 0 nbsp; amp; 0 nbsp; 1 amp; 1 amp; 0 ...

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Standard XII

Mathematics

Perpendicular Distance of a Point from a Plane

OABC is a tet...

Question

$O A B C$ is a tetrahedron. The position vectors of $A, B$ and $C$ are $^i,^i +^j,^j +^k$ respectively. $^{'} O^{'}$ is the origin. The distance of the plane face $A B C$ from origin $O$ is

\frac{1}{2}

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\frac{1}{\sqrt{2}}

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\frac{1}{2 \sqrt{2}}

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Solution

The correct option is A $\frac{1}{\sqrt{2}}$
Volume of tetrahedron
$= \frac{1}{6} [\begin{matrix} - - \to O A - - \to O B - - \to O C \end{matrix}] = \frac{1}{6} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 0 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = \frac{1}{6}$
Area of face $A B C, = \frac{1}{2} ∣ ∣ \begin{matrix} (\begin{matrix} ^i +^j -^i \end{matrix}) \times (\begin{matrix} ^j +^k -^i \end{matrix}) \end{matrix} ∣ ∣$
$= \frac{1}{2} ∣ ∣ \begin{matrix} ^i +^k \end{matrix} ∣ ∣ = \frac{1}{\sqrt{2}}$
$∴$ Required distance
$= \frac{3 \times v o l u m e}{A r e a o f b a s e} = \frac{3 \sqrt{2}}{6} = \frac{1}{\sqrt{2}}$

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Similar questions

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OABC is a tetrahedron. The position vectors of A,B and C are ^i,^i+^j,^j+^k respectively. ′O′ is the origin. The distance of the plane face ABC from origin O is

The correct option is A 1√2Volume of tetrahedron=16[−−→OA−−→OB−−→OC]=16∣∣ ∣∣100110011∣∣ ∣∣=16Area of face ABC,=12∣∣(^i+^j−^i)×(^j+^k−^i)∣∣=12∣∣^i+^k∣∣=1√2∴ Required distance=3×volumeAreaofbase=3√26=1√2

$O A B C$ is a tetrahedron. The position vectors of $A, B$ and $C$ are $^i,^i +^j,^j +^k$ respectively. $^{'} O^{'}$ is the origin. The distance of the plane face $A B C$ from origin $O$ is