In a tetrahedron LMNO,edges ML, MN and MO are | vedclass

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(A) Let the vertex $M$ be at the origin $(0, 0, 0)$. Let the lengths of the edges $ML, MN, MO$ be $a, b, c$ respectively.The equation of the plane pas

Vedclass · Accepted Answer

$\frac{6}{7} 	ext{ units}$

Vedclass · Answer

(A) Let the vertex $M$ be at the origin $(0, 0, 0)$. Let the lengths of the edges $ML, MN, MO$ be $a, b, c$ respectively.The equation of the plane passing through $L(a, 0, 0), N(0, b, 0)$ and $O(0, 0, c)$ is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.The altitude from $O$ to face $MLN$ is $c = 1$. The altitude from $L$ to face $MNO$ is $a = 2$. The altitude from $N$ to face $MLO$ is $b = 3$.Thus,the equation of the face $LNO$ is $\frac{x}{2} + \frac{y}{3} + \frac{z}{1} = 1$,which simplifies to $\frac{x}{2} + \frac{y}{3} + z - 1 = 0$.The length of the altitude $h$ from $M(0, 0, 0)$ to the face $LNO$ is given by the distance formula:$h = \frac{|-1|}{\sqrt{(\frac{1}{2})^2 + (\frac{1}{3})^2 + (1)^2}} = \frac{1}{\sqrt{\frac{1}{4} + \frac{1}{9} + 1}} = \frac{1}{\sqrt{\frac{9+4+36}{36}}} = \frac{1}{\sqrt{\frac{49}{36}}} = \frac{6}{7} 	ext{ units}$.

In a tetrahedron $LMNO$,edges $ML, MN$ and $MO$ are mutually perpendicular. If the lengths of the altitudes drawn from $O, L$ and $N$ to their opposite faces are $1, 2$ and $3$ units respectively,then the length of the altitude drawn from $M$ to the face $LNO$ is:

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