Question

Find the coordinates of the foot of the perpendicular from the point (2,3,-8) to the line (4-x)/2 = y/6 = (1-z)/3. Also find the perpendicular distance from the given point to the given line.

Answer 1

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Find the foot of perpendicular from the point(2,3,-8)to the line 4−x2=y6=1−z34−x2=y6=1−z3.Also,find the perpendicular distance from the given point to the line.

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asked Jan 3, 2013 by sreemathi.v 
retagged May 23, 2013 by rvidyagovindarajan_1



1 Answer

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If two lines are ⊥⊥ then a1a2+b1b2+c1c2=0a1a2+b1b2+c1c2=0, where (a1,b1,c1)(a1,b1,c1)and (a2,b2,c2)(a2,b2,c2) are the direction ratios of the two lines.

Step 1:

Let the given equation be 4−x2=y6=1−z3=λ4−x2=y6=1−z3=λ.

This can be written as

x−4−2=y−06=z−1−3x−4−2=y−06=z−1−3=λ=λ-----(1)

Therefore the direction ratios of the line is(−2,6,−3)(−2,6,−3)

Therefore the coordinates of any point on the line is

x=4−2λ,y=6λ,z=1−3λx=4−2λ,y=6λ,z=1−3λ

Step 2:

Let Q(4−2λ,6λ,1−3λ)Q(4−2λ,6λ,1−3λ) be the foot of ⊥⊥ from the point P(2,3,−8)P(2,3,−8) on line (1)

We know the direction ratios of any line segement PQPQ is given by (x2−x1),(y2−y1),(z2−z1)(x2−x1),(y2−y1),(z2−z1)

The direction cosines of PQ¯¯¯¯¯¯¯¯PQ¯ is given by

=(−2λ+4−2),(6λ−3),(−3λ+1+8)=(−2λ+4−2),(6λ−3),(−3λ+1+8)

=(−2λ+2,6λ−3,−3λ+9)=(−2λ+2,6λ−3,−3λ+9)

Now QQ is the foot of the perpendicular of the line (1)

(ie) PQ−→−PQ→ is the perpendicular to the line (i)

hence the sum of the product of this direction ratios is 0

=(−2λ+2).(−2)+(6λ−3).6+(−3λ+9).(−3)=0=(−2λ+2).(−2)+(6λ−3).6+(−3λ+9).(−3)=0

=4λ−4+36λ−18+9λ−27=0=4λ−4+36λ−18+9λ−27=0

On simplifying we get,

49λ−49=049λ−49=0

Therefore λ=1λ=1

Step 3:

Substituting λλ in QQ we get the

Q(0,3,6)Q(0,3,6)

Therefore the perpendicular distance =02+32+62−−−−−−−−−−√=02+32+62

=45−−√