Question

If k is the length of any edge of a regular tetrahedron then the distance of any vertex from the opposite face is –

• A. 3/2 k
• B. 2/3 k²
• C. $\sqrt{\frac{2}{3}}$k
• D. $\sqrt{3}$k

Answer 1

Answer: (C)

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

Answer 2

OABC be a regular tetrahedron.. O be the origin Position vector of ABC be a, b, c; |a| = |b| = |c| = k

|a.b| = |b.c| = |c.a| = k² cos 60º =$\frac{1}{2}$k²

a.a = b.b = c.c = k²

[a, b, c] = $\left| \begin{matrix} a.a & a.b & a.c \\ b.a & b.b & b.c \\ c.a & c.b & c.c \end{matrix} \right|$ ̃ $\frac{1}{2}$ k⁶

Also |a × c + c × a + a × b| is twice the area of the triangle ABC so this is ($\sqrt{3}$/2)k²

The equation of the plane ABC is

r. [b × c + c × a + a × b] ̃ [a b c]

So the distance of the vector O, from this plane

$\frac{\lbrack abc\rbrack}{\lbrack b \times c + c \times a + a \times b\rbrack}$ ̃ $\frac{\frac{1}{\sqrt{2}}k^{3}}{\frac{\sqrt{3}}{2}k^{2}}$ ̃ $\sqrt{\frac{2}{3}}$k