The coordinates of the vertices P,Q and R of a triangle PQR are (1,-1,1), (3,-2,2) and (0,2,6) respectively. If `angleRQP=theta`, then what is `angle PRQ` equal to ?

Options

A. `30^@=theta`
B. `45^@+theta`
C. `60^@-theta`
D. `90^@-theta`

Answer

D

Explanation

Given,
`P=(1,-1,1)`
`Q=(3,-2,2)`
`R=(0,2,6)`

Direction ratios of PQ`(a_1,b_1,c_1)=(3-1,-2+1,2-1) =(2,-1,1)`
Direction ratios of PR `(a_2,b_2,c_2)=(0-1,2+1,6-1)=(-1,3,5)`
Let us calculate, `a_1a_2+b_1b_2+c_1c_2`
`=(2)(-1)+(-1)(3)+(1)(5)`
`=-2-3+5`
=0.
`:.PQ_|_ PR i.e., angle QPR=90^@`
In `angle PQR,angle P+angleQ+angle R=180^@rArr 90^@+theta+angleR=180^@`
`rArr angleR=90^@-theta`