The vertices of a triangle are A(1, 1, 1), B(3, 2, 3) and C(2, 1, 0).
(A) The line AB is perpendicular to the line AC
(B) The direction cosines of AB are \(\displaystyle\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\)
(C) The line BC is parallel to the line \(\displaystyle\frac{x-1}{1}=\frac{y+2}{1}=\frac{z}{3}\)
(D) The point (5, 3, 4) lies on the line AB
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (C) only
The direction ratios of the sides are the differences of coordinates:
AB: 2, 1, 2
AC: 1, 0, −1
BC: −1, −1, −3
\((2)(1)+(1)(0)+(2)(-1) = 0\)
\(\sqrt{4+1+4} = 3\)
The general point of AB is \((1+2t,\ 1+t,\ 1+2t)\). For x = 5 we need t = 2, which gives (5, 3, 5).
So AB and AC have dot product 0, AB has length factor 3, BC has direction −1, −1, −3, and the point of AB with x = 5 is (5, 3, 5).
Checking the statements
(A) AB is perpendicular to AC — Correct. \(2+0-2 = 0\).
(B) The direction cosines of AB are \(\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\) — Correct. They are 2, 1, 2 divided by 3.
(C) BC is parallel to the line with direction ratios 1, 1, 3 — Correct. −1, −1, −3 is −1 times 1, 1, 3.
(D) The point (5, 3, 4) lies on AB — Incorrect. The point of AB with x = 5 is (5, 3, 5).