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
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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
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