Question

The coordinates of the vertices of a tetrahedron ABCD are as follows A (2, 3,4); B (1, –1, 2); C (0, 4, 5); D(–2, 3, –4). Then, which of the following is true–

• A. the angle between lines AB and CD is$\left( \frac{24}{\sqrt{1860}} \right)$
• B. the angle between AD and the plane ABC is$\sqrt{\frac{80}{11}}$ sin^–1$\sqrt{\frac{80}{11}}$
• C. the equation of line BD is$\frac{x - 1}{3}$=$\frac{y - 1}{- 4}$=$\frac{z - 2}{6}$
• D. the perpendicular distance from D to the plane ABC is $\frac{80}{\sqrt{110}}$

Answer 1

Answer: (D)

the perpendicular distance from D to the plane

ABC is $\frac{80}{\sqrt{110}}$

Answer 2

Dr's of AB 1, 4, 2 and of CD 2, 1, 9

angle between AB and CD

= cos^–1 $\left( \frac{1.2 + 4.1 + 2.9}{\sqrt{1^{2} + 4^{2} + 2^{2}}\sqrt{2^{2} + 1^{2} + 9^{2}}} \right)$ = cos^–1 $\left( \frac{24}{\sqrt{1806}} \right)$

equation of plane a (x –2) + b(y – 3) + c(z – 4) = 0

Hence (1) is not true.

It passes through B and C Ž a = 2k, b = –5k, c = 9k

Ž 2x – 5y + 9z – 25 = 0

\ dr's of plane ABC is 2, –5, 9

dr's of line AB is 4, 0, 8

angle between AD and plane ABC = sin^–1$\sqrt{\frac{8}{11}}$

Hence (2) is not true.

dr's of BD is 3, –4, 6 so BD $\frac{x - 1}{3}$= $\frac{y + 1}{- 4}$= $\frac{z - 1}{6}$

Hence (3) is not true.

perpendicular distance from D to ABC

= $\left| \frac{2( - 2) - 5(3) + 9( - 4) - 25}{\sqrt{2^{2} + ( - 5)^{2} + 9^{2}}} \right|$

= $\frac{80}{\sqrt{110}}$Hence (4) is true.