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Question: Volume of the tetrahedron PQRS (in cu. units) is -...

Volume of the tetrahedron PQRS (in cu. units) is -

A

472\frac{47}{2}

B

20

C

25

D

572\frac{57}{2}

Answer

20

Explanation

Solution

Let the coordinate system be set up such that P is at the origin (0,0,0) and Q is on the z-axis. Since PQ = 4, Q is at (0,0,4). The vector PQ⃗=(0,0,4)\vec{PQ} = (0,0,4).

The plane P1P_1 is perpendicular to PQ. Thus, the normal vector to P1P_1 is parallel to PQ⃗\vec{PQ}. Let the normal vector be (0,0,1)(0,0,1). The equation of the plane P1P_1 is of the form 0x+0y+1z=d0x + 0y + 1z = d, or z=dz=d for some constant dd.

P', R', S' are the feet of the perpendiculars drawn from P, R, S to the plane P1P_1. Let R = (xR,yR,zR)(x_R, y_R, z_R) and S = (xS,yS,zS)(x_S, y_S, z_S). The line from P(0,0,0) perpendicular to z=dz=d is (0,0,0)+t(0,0,1)=(0,0,t)(0,0,0) + t(0,0,1) = (0,0,t). This line intersects z=dz=d when t=dt=d. So P' = (0,0,d). The line from R(xR,yR,zR)(x_R, y_R, z_R) perpendicular to z=dz=d is (xR,yR,zR)+t(0,0,1)=(xR,yR,zR+t)(x_R, y_R, z_R) + t(0,0,1) = (x_R, y_R, z_R+t). This line intersects z=dz=d when zR+t=dz_R+t=d, so t=d−zRt=d-z_R. So R' = (xR,yR,zR+(d−zR))=(xR,yR,d)(x_R, y_R, z_R + (d-z_R)) = (x_R, y_R, d). Similarly, S' = (xS,yS,d)(x_S, y_S, d).

The points P', R', S' lie in the plane z=dz=d. P′R′⃗=(xR−0,yR−0,d−d)=(xR,yR,0)\vec{P'R'} = (x_R - 0, y_R - 0, d - d) = (x_R, y_R, 0). P′S′⃗=(xS−0,yS−0,d−d)=(xS,yS,0)\vec{P'S'} = (x_S - 0, y_S - 0, d - d) = (x_S, y_S, 0). The area of △P′R′S′\triangle P'R'S' is given as 15 sq. units. Area(△P′R′S′\triangle P'R'S') = 12∣P′R′⃗×P′S′⃗∣\frac{1}{2} |\vec{P'R'} \times \vec{P'S'}|. P′R′⃗×P′S′⃗=∣ijkxRyR0xSyS0∣=(xRyS−yRxS)k\vec{P'R'} \times \vec{P'S'} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_R & y_R & 0 \\ x_S & y_S & 0 \end{vmatrix} = (x_R y_S - y_R x_S)\mathbf{k}. Area(△P′R′S′\triangle P'R'S') = 12∣xRyS−yRxS∣=15\frac{1}{2} |x_R y_S - y_R x_S| = 15. So, ∣xRyS−yRxS∣=30|x_R y_S - y_R x_S| = 30.

Volume of the tetrahedron PQRS. The volume of a tetrahedron with vertices P, Q, R, S is given by V=16∣det⁡(PQ⃗,PR⃗,PS⃗)∣V = \frac{1}{6} |\det(\vec{PQ}, \vec{PR}, \vec{PS})|. With P=(0,0,0), Q=(0,0,4), R=(xR,yR,zR)(x_R, y_R, z_R), S=(xS,yS,zS)(x_S, y_S, z_S): PQ⃗=(0,0,4)\vec{PQ} = (0,0,4). PR⃗=(xR,yR,zR)\vec{PR} = (x_R, y_R, z_R). PS⃗=(xS,yS,zS)\vec{PS} = (x_S, y_S, z_S). V=16∣∣004xRyRzRxSySzS∣∣=16∣0(yRzS−zRyS)−0(xRzS−zRxS)+4(xRyS−yRxS)∣V = \frac{1}{6} \left| \begin{vmatrix} 0 & 0 & 4 \\ x_R & y_R & z_R \\ x_S & y_S & z_S \end{vmatrix} \right| = \frac{1}{6} |0(y_R z_S - z_R y_S) - 0(x_R z_S - z_R x_S) + 4(x_R y_S - y_R x_S)| V=16∣4(xRyS−yRxS)∣V = \frac{1}{6} |4 (x_R y_S - y_R x_S)|. Using ∣xRyS−yRxS∣=30|x_R y_S - y_R x_S| = 30, we get: V=16∣4×30∣=1206=20V = \frac{1}{6} |4 \times 30| = \frac{120}{6} = 20. The volume of the tetrahedron PQRS is 20 cu. units.