Question

F Q (2,1,-2) and R(0,-5,1)then the perpendicular distance from
p(1,4,-2) to (QR) is​

Answer 1

Answer:

QR makes equal angles with axis ±

3

1

DR of QR are (±

3

1

,±

3

1

,±

3

1

)

DR of QR are (±1,±1,±1)

⇒QR:

1

x+3

=

1

y−6

=

1

z−2

=t

Foot of perpendicular from P be P

′

(t−3,t+6,t+2)

DR's of PP

′

are perpendicular to QR

DR's of PP

′

(t−3−1−2,t+6−3,t+2−1)

=(t−1,t+3,t+1)

PP

′

.

QR

=0

⇒(t−1)+t+3+t+1=0

3t+3=0

t=−1

⇒P

′

(−4,5,1)

∣PP

′

∣=

(−2+4)

2

+(3−5)

2

+(1−1)

2

=2

2