Question

Find the coordinates of the foot of the perpendicular from the point ( , , ) 0 2 3 on the line x y z + = = +3 5 1 2 4 3 . Also find the length of the perpendicular

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Find the coordinates of the foot of the perpendicular drawn from the point (0,2,3)(0,2,3) on the line \(\large\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3} \). Also, find length of perpendicular.

cbse

class12

modelpaper

2012

sec-c

q27

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math

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asked Feb 10, 2013 by thanvigandhi_1 
edited Sep 27, 2013 by sreemathi.v



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Length of the ⊥⊥ from point on the given line=(x2−x1)2+(y2−y1)2+(z2−z1)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−√(x2−x1)2+(y2−y1)2+(z2−z1)2

Let LL be the foot of the ⊥⊥ drawn from the point P(0,2,3)P(0,2,3) to the given line.

The coordinates of a general point on x+35=y−12=z+43x+35=y−12=z+43 are given by

x+35=y−12=z+43=x+35=y−12=z+43=λλ

(i.e) x=5λ−3x=5λ−3----(1)

y=2λ+1y=2λ+1-----(1)

z=3λ−4z=3λ−4------(1)

This is represented by LL in the figure.



Step 2:

∴∴ Direction ratios of PL are proportional to 5λ−3−0,2λ−1−2,3λ−4−35λ−3−0,2λ−1−2,3λ−4−3

(i.e) 5λ−3,2λ−1,3λ−75λ−3,2λ−1,3λ−7

The direction ratios of the given line are proportional to 5,2,35,2,3

Since PLPL is ⊥⊥ to the given line

a1a2+b1b2+c1c2=0a1a2+b1b2+c1c2=0

5(5λ−3)+2(2λ−1)+3(3λ−7)=05(5λ−3)+2(2λ−1)+3(3λ−7)=0

On simplifying we get,

λ=1λ=1

Step 3:

Put λ=1λ=1 in equ(1) we get

x=2,y=3x=2,y=3 and z=−1z=−1

∴∴ Length of the ⊥⊥ from PP on the given line is

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

=(2−0)2+(3−2)2+(−1−3)2−−−−−−−−−−−−−−−−−−−−−−−−−√=(2−0)2+(3−2)2+(−1−3)2

=21−−√=21units