For the two
lines r1(t) = (2,4,-1) +t(0,0,2) and
r2(s) = (13,-1,o] +s (0,-3,0), the shortest
distance between the two lines is what??
Answers
Step-by-step explanation:
Comparing the line
r
=(1−t)
i
^
+(t−2)
j
^
+(3−2t)
k
^
or
r
=
i
^
−2
j
^
+3
j
^
+3
k
^
+t(−
i
^
+
j
^
−2
k
^
)
with
r
=
a
1
+λ
b
1
a
1
=
i
^
−2
j
^
+3
k
^
,
b
1
=−
i
^
+
j
^
−2
k
^
,
and comparing line
r
=(s+1)
i
^
+(2s−1)
j
^
−(2s+1)
k
^
r
=
i
^
−
j
^
−
k
^
+s(
i
^
+2
j
^
−2
k
^
)
with
r
=
a
2
+μ
b
2
a
2
=
i
^
−
j
^
−
k
^
,
b
2
=
i
^
+2
j
^
−2
k
^
∴
r
=
a
1
+λ
b
1
and
r
=
a
2
+μ
b
2
∴ Distance between
r
=
a
1
+λ
b
1
and
r
=
a
2
+μ
b
2
d=
∣
∣
∣
∣
∣
∣
∣
b
1
×
b
2
∣
(
a
2
−
a
1
).(
b
1
×
b
2
)
∣
∣
∣
∣
∣
∣
....(1)
∴
a
2
−
a
1
=(
i
^
−
j
^
−
k
^
)−(
i
^
−2
j
^
+3
k
^
)
=0+
j
^
−4
k
^
=
j
^
−4
k
^
and
b
1
×
b
2
=(−
i
^
+
j
^
−2
k
^
)×(
i
^
+2
j
^
−2
k
^
)
=
∣
∣
∣
∣
∣
∣
∣
∣
i
^
−1
1
j
^
1
2
k
^
−2
−2
∣
∣
∣
∣
∣
∣
∣
∣
=(−2+4)
i
^
−(2+2)
j
^
+(−2−1)
k
^
=2
i
^
−4
j
^
−3
k
^
∴∣
b
1
×
b
2
∣=
(2)
2
+(−4)
2
+(−3)
2
=
171
=
29
∴d=
∣
∣
∣
∣
∣
∣
29
(
j
^
−4
k
^
).(2
i
^
−4
j
^
−3
k
^
)
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
29
−4+12
∣
∣
∣
∣
∣
=
29
8