Question

If l is the length of any edge of a regular tetrahedron, then the distance of any vertex from the opposite face is-

• A. $\frac { 2 } { 3 }$ l²
• B. $\sqrt { \frac { 2 } { 3 } }$l
• C. $\frac { \sqrt { 2 } } { 3 }$l
• D. None of these

Answer 1

Answer: (B)

$\sqrt { \frac { 2 } { 3 } }$l

Answer 2

Let OABC be a regular tetrahedron such that

$\overrightarrow { \mathrm { OA } }$= , $\overrightarrow { \mathrm { OB } }$= , $\overrightarrow { \mathrm { OC } }$= and ||= || = || = l.

Now |.| = | . | = |. | = l² cos 60° = $\frac { \lambda ^ { 2 } } { 2 }$

and |.| = |.| = |.| = l²

[] = = $\frac { \lambda ^ { 2 } } { 2 }$

Also |×+ × + × | = 2 Area of DABC

i.e. $\frac { \sqrt { 3 } } { 2 }$ l²

Also equation of plane ABC is

$\overrightarrow { \mathrm { r } }$ . [×+ $\overrightarrow { \mathrm { b } }$ × $\overrightarrow { \mathrm { c } }$ + $\overrightarrow { \mathrm { c } }$ ×] = [ $\overrightarrow { \mathrm { c } }$ ]

\ distance of O from plane

= $\frac { [ \vec { a } \vec { b } \vec { c } ] } { [ \vec { a } \times \vec { b } + \vec { b } \times \vec { c } + \vec { c } \times \vec { a } ] }$ = $\frac { \lambda ^ { 3 } / \sqrt { 2 } } { \frac { \sqrt { 3 } \lambda ^ { 2 } } { 2 } }$ = l $\sqrt { \frac { 2 } { 3 } }$