The shortest distance between the diagonals of a rectangular parallelopiped whose sides are and the edges not meeting it are:
Answers
Step-by-step explanation:
Take O as the origin of vectors and let i,j,k denote unit vectors along
OA
,
OB
and
OC
respectively.
Then,
OA
=ai,
OB
=bj,
OC
=ck
Also
OP
= ai+bj+ck. The edges which do not meet the diagonal OP and BL,BN,AN and this parallels AM,CMandCL .
Suppose we are to find the distance between the diagonal OP and the edge AN.
Now, OP is the line passing through O whose position vector is
O
and parallel to the vector ai+bj+ck.
And AN is the line through A whose position vector is ai and paraallel to j. Hence, the shortest distance between OP and AN is given by P=
∣j×(ai+bj+ck)∣
[
O
−ai,j,ai+bj+ck]
Now, [
O
−ai,j,ai+bj+ck] =
∣
∣
∣
∣
∣
∣
∣
∣
−a
0
a
0
1
b
0
0
0
∣
∣
∣
∣
∣
∣
∣
∣
=ca
and j×(ai+bj+ck)=
∣
∣
∣
∣
∣
∣
∣
∣
i
0
a
j
1
b
k
0
0
∣
∣
∣
∣
∣
∣
∣
∣
=ci−ak
So that ∣j×(ai+bj+ck)∣=
c
2
+a
2
∴P=
c
2
+a
2
ca
Similarly, it can be shown that the shortest distance between OP and BN is
c
2
+a
2
ca
and that between OP and BL is
a
2
+b
2
ab
solution
December 20, 2019
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Practice important Questions
Vectors and 3D Geometry II
231 Qs
Solve related Questions
find the shortest distance between the lines
−1
x−2
=
2
y−5
=
3
z−0
and
2
x−0
=
−1
y+1
=
2
z−1
1 Verified Answer
If the shortest distance between the lines x+2λ=2y=−12z,x=y+4λ=6z−12λ is 4
2
MAKE ME BRILLIANT....
FOLLOW ME...
Step-by-step explanation:
Take O as the origin of vectors and let i,j,k denote unit vectors along
OA
,
OB
and
OC
respectively.
Then,
OA
=ai,
OB
=bj,
OC
=ck
Also
OP
= ai+bj+ck. The edges which do not meet the diagonal OP and BL,BN,AN and this parallels AM,CMandCL .
Suppose we are to find the distance between the diagonal OP and the edge AN.
Now, OP is the line passing through O whose position vector is
O
and parallel to the vector ai+bj+ck.
And AN is the line through A whose position vector is ai and paraallel to j. Hence, the shortest distance between OP and AN is given by P=
∣j×(ai+bj+ck)∣
[
O
−ai,j,ai+bj+ck]
Now, [
O
−ai,j,ai+bj+ck] =
∣
∣
∣
∣
∣
∣
∣
∣
−a
0
a
0
1
b
0
0
0
∣
∣
∣
∣
∣
∣
∣
∣
=ca
and j×(ai+bj+ck)=
∣
∣
∣
∣
∣
∣
∣
∣
i
0
a
j
1
b
k
0
0
∣
∣
∣
∣
∣
∣
∣
∣
=ci−ak
So that ∣j×(ai+bj+ck)∣=
c
2
+a
2
∴P=
c
2
+a
2
ca
Similarly, it can be shown that the shortest distance between OP and BN is
c
2
+a
2
ca
and that between OP and BL is
a
2
+b
2
ab
solution
December 20, 2019
avatar
Toppr
SHARE
Practice important Questions
Vectors and 3D Geometry II
231 Qs
Solve related Questions
find the shortest distance between the lines
−1
x−2
=
2
y−5
=
3
z−0
and
2
x−0
=
−1
y+1
=
2
z−1
1 Verified Answer
If the shortest distance between the lines x+2λ=2y=−12z,x=y+4λ=6z−12λ is 4
2
MAKE ME BRILLIANT....
FOLLOW ME...
Thanks!!!!!