Altitude in tetrahedron I was reading a book there , I saw the below question in the book  . 

My doubt is that how they have written length of altitude is same as projection of AD on ABC  ?

Please explain if possible with a diagram
Hope the following picture can clarify ……
The idea is that volume of the tetrahedron equals $1/3$ base $\times$ height.

Or, Height = 3V / Area of the base.

But rather than working with tetrahedra and triangles, the math is a little nicer if we use parallelepeiped and parallelograms.

In which case, Height = Volume / Area.

Translate everything such that $A$ is at the origin.

The volume of the parallelepeiped formed by the vectors $AB, AC, AD$ equals the absolute value of the determinant of the matrix with those vectors as rows.

The area of parallelogram formed by $AB$ and $AC$ equals the absolute value of cross product $AB\times AC$