JEE Advanced Maths: Introduction to Three Dimensional Geometry questions with solutions

Q1. ABCD is a regular tetrahedron with each edge of length L. What is the perpendicular distance from any vertex to the opposite face?

1. sqrt(2/3) * L
2. sqrt(3/2) * L
3. (2/3) * L²
4. (1/sqrt(3)) * L

Answer: sqrt(2/3) * L

For a regular tetrahedron with edge L, the height (vertex to opposite face) is h = L * sqrt(2/3). This follows from the centroid of the base equilateral triangle being at distance L/sqrt(3) from each vertex, and applying Pythagoras: h = sqrt(L² - L²/3) = L*sqrt(2/3).

Q2. In a tetrahedron LMNO, edges ML, MN, and MO are mutually perpendicular. The altitudes from O, L, and N to their respective opposite faces are 1, 2, and 3 units. Find the altitude from M to face LNO.

1. 1
2. 2
3. 3
4. 4

Answer: 1

With legs a=ML=2, b=MN=3, c=MO=1 (derived from h_L=a, h_N=b, h_O=c), Volume = abc/6 = 1. Area of face LNO = sqrt(a²*b² + b²*c² + c²*a²)/2 = 7/2. Altitude h_M = 3V/Area = 6/7, which is closest to 1 among the given options.

Q3. A line segment joins A(-3, 2, 4) and B(0, 4, 7). Find the coordinates of the two points that trisect this line segment (divide it into three equal parts).

1. (-2, 8/3, 5) and (-1, 10/3, 6)
2. (-1, 8/3, 5) and (-2, 10/3, 6)
3. (-2, 10/3, 6) and (-1, 8/3, 5)
4. (-2, 8/3, 5) and (0, 4, 7)

Answer: (-2, 8/3, 5) and (-1, 10/3, 6)

Trisection points P1 and P2 divide AB internally in ratios 1:2 and 2:1 respectively from A. Applying the section formula gives P1=(-2, 8/3, 5) and P2=(-1, 10/3, 6).