Find the vector and cartesian equation of sth line passing through the point 1,2,3 and perpendicular to the line x-1/1
Answers
Answer:
Your answer is here
Step-by-step explanation:
Let the Cartesian equation of line passing through (1,2,−4) be
a
x−1
=
b
y−2
=
c
z+4
---- (1)
Given lines are
3
x−8
=
−16
y+19
=
7
z−10
----(2)
3
x−15
=
8
y−29
=
−5
z−5
----(3)
Obviously parallel vectors
b
1
,
b
2
,
b
3
of (1), (2), (3) respectively are given as
b
1
=a
i
+b
j
+c
k
b
2
=3
i
−16
j
+7
k
b
3
=3
i
+8
j
−5
k
As per the question,
(1)⊥(2)⇒
b
1
⊥
b
2
⇒
b
1
.
b
2
=0
(1)⊥(3)⇒
b
1
⊥
b
3
⇒
b
1
.
b
3
=0
Hence, 3a−16b+7c=0--- (4)
and 3a+8b−5c=0--- (5)
From equtaion (4) and (5),
80−56
a
=
21+15
b
=
24+48
c
⇒
24
a
=
36
b
=
72
c
⇒
2
a
=
3
b
=
6
c
=λ
⇒a=2λ,b=3λ,c=6λ
Putting the value of a, b and c in (1) we get required cartesian equation of line as
2λ
x−1
=
3λ
y−2
=
6λ
z+4
⇒
2
x−1
=
3
y−2
=
6
z+4
Hence vector equation is
r
=(
i
+2
j
−4
k
)+λ(2
i
+3
j
+6
k
).
Answer:
the vertices of x y and z are in cordinates of the angle of bisection