Question

Find the length and foot of the perpendicular drawn from a point 2,-1,5

Answer 1

Let L be the foot of the perpendicular drawn from the point P(2,−1,5) to the given line.

The coordinates of a general point on x−1110=y+2−4=z+8−11 are given by

x−1110=y+2−4=z+8−11=k

x=10k+11,y=−4k−2,z=−11k−8

Let the coordinates of L be (10k+11,−4k−2,−11k−8)

∴ Direction ratios of PL are proportional to 10k+11−2,−4k−2+1,−11k−8−5

(ie)10k+9,−4k−1,−11k−13

Direction ratios of the given line are proportional to (10,−4,−11)

Step 2:

Since PL is ⊥ to the given line

∴10(10k+9)−4(−4k−1)−11(−11k−13)=0

On simplifying we get,

90+100k+16k+4+121k+143=0

237k=−237

⇒k=−1

Hence the coordinates of L are (1,2,3)

PL=(x2−x1)2+(y2−y1)2+(z2−z1)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

=(2−1)2+(−1−2)2+(5−3)2−−−−−−−−−−−−−−−−−−−−−−−−−√

=12+(−3)2+(2)2−−−−−−−−−−−−−−√

=√14units.