16. Find the angle between the two lines whose direction cosines are given by the relations.
i) l + m + n = 0 and l² + m² -n² = 0
Answers
Answer:
hope this helps u
Step-by-step explanation:
Let l1m1n1 and l2m2n2 be the direction cosines.
l+m+m=0⇒l=−m−nal2+bm2+cn2=0a(m2+n2+2mn)+bm2+cn2=0(a+b)m2+(a+c)n2+2amn=0
(a+b)(nm)2+2a(nm)+(a+c)=0 -Quadratic in nm
∴ product of roots=n1m1×n2m2=a+
Answer:
hope it helps you mate
Step-by-step explanation:
Let l
1
m
1
n
1
and l
2
m
2
n
2
be the direction cosines.
l+m+m=0⇒l=−m−n
al
2
+bm
2
+cn
2
=0
a(m
2
+n
2
+2mn)+bm
2
+cn
2
=0
(a+b)m
2
+(a+c)n
2
+2amn=0
(a+b)(
n
m
)
2
+2a(
n
m
)+(a+c)=0 -Quadratic in
n
m
∴ product of roots=
n
1
m
1
×
n
2
m
2
=
a+b
a+c
Similarly,
m=−l−n
al
2
+b(l
2
+n
2
+2ln)+cn
2
=0
(a+b)l
2
+(b+c)n
2
+2bln=0
(a+b)(
n
l
)
2
+2b(
n
l
)+(b+c)=0− Quadratic in
n
l
Product of roots =
n
1
l
1
×
n
2
l
2
=
a+b
b+c
Since the two lines are perpendicular,
∴l
1
l
2
+m
1
m
2
+n
1
n
2
=cos90
l
1
l
2
+m
1
m
2
+n
1
n
2
=0
n
1
n
2
l
1
l
2
+
n
1
n
2
m
1
m
2
+1=
n
1
n
2
0
a+b
a+c
+
a+b
b+c
+1=0
2(a+b+c)=0
a+b+c=0