Question

Find the equation of the straight line perpendicular to 5 − = 1 and passing through

the point (2,3) and also find their point of intersection​

Answer 1

Answer:

Let a,b,c are Dr

′

s of required line, thus equation of line passing through (1,2,3)

a

x−1

=

b

y−2

=

c

z−3

=k(say) ...(1)

⇒(ak+1,bk+2,ck+3) is any general point on (1).

Also, given line that intersect (1) is

2

x−(−1)

=

1

y−2

=

2

z−(−4)

=λ ....(2)

⇒(2λ−1,λ+2,2λ−4) is any general point on (2)

∵ line (1) and (2) are intersecting

a=

k

2λ−2

;b=

k

λ

;c=

k

2λ−7

∵ (1) is parallel to plane x+5y+4z=0 i.e. perpendicular to normal vector.

So a+5b+4c=0

⇒(

k

2λ−2

)+5

k

λ

+4(

k

2λ−7

)=0⇒λ=2

⇒a=

k

2

;b=

k

2

;c=−

k

3

Therefore required line is

2

x−1

=

2

y−2

=

−3

z−3