wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If foot of the perpendicular of P(2,-3,1) on the line
x+12=y33=z+21is
Q(a,b,c) then find the value of 14(a+b+c)

Open in App
Solution

Given line is x+12=y33=z+21=r...(1)
Point P(2,3,1)
Co-ordinates of foot of the perpendicular on line (1) may be taken as Q(2r1,3r+3,r2)
We get direction ratios of PQ=2r3,3r+6,r3
Direction ratios of line segment are 2,3,1 from (1)
Since PQ perpendicular to AB
2(2r3)+3(3r+6)1(r3)=0
or,14r+15=0
r=1514
Q(2r1,3r+3,r2)=(227,314,1314)
=(a,b,c) given
14(a+b+c)=14(227+1314+1314)=44+3+13=60


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Foot of Perpendicular, Image and Angle Bisector
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon