Let OABC be a tetrahedron O being the origin- If position vectors of A - B and C are - - and -k- respectively- then height of the tetrahedron taking ABC as base is equal to A- 1-2B- 1-2C- 1-2 -2D- -2

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

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Standard XII

Mathematics

Perpendicular Distance of a Point from a Plane

Let OABC be a...

Question

Let OABC be a tetrahedron (O being the origin). If position vectors of A, B and C are $^i,^i +^j$ and $^j +^k$ respectively, then height of the tetrahedron (taking ABC as base) is equal to

\frac{1}{2}

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

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

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\sqrt{2}

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Solution

The correct option is B $\frac{1}{\sqrt{2}}$
The volume of tetrahedron $= \frac{1}{6} | [¯ ¯¯¯¯¯¯ ¯ O A ¯ ¯¯¯¯¯¯ ¯ O B ¯ ¯¯¯¯¯¯ ¯ O C] | = \frac{1}{6} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 0 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = \frac{1}{6}$
Area of base $= \frac{1}{2} | (^i +^j -^i) \times (^j +^k -^i) | = \frac{1}{2} |^i \times^k | = \frac{1}{2} (\sqrt{2}) = \frac{1}{\sqrt{2}}$
Hence height $= \frac{3 \times v o l u m e}{A r e a o f b a s e} = \frac{3 \sqrt{2}}{6} = \frac{1}{\sqrt{2}}$

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Similar questions

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Let OABC be a tetrahedron (O being the origin). If position vectors of A, B and C are ^i,^i+^j and ^j+^k respectively, then height of the tetrahedron (taking ABC as base) is equal to

The correct option is B 1√2The volume of tetrahedron =16|[¯¯¯¯¯¯¯¯OA ¯¯¯¯¯¯¯¯OB ¯¯¯¯¯¯¯¯OC]|=16∣∣ ∣∣100110011∣∣ ∣∣=16 Area of base =12|(^i+^j−^i)×(^j+^k−^i)|=12|^i×^k|=12(√2)=1√2 Hence height =3×volumeArea of base=3√26=1√2

Let OABC be a tetrahedron (O being the origin). If position vectors of A, B and C are $^i,^i +^j$ and $^j +^k$ respectively, then height of the tetrahedron (taking ABC as base) is equal to