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Byju's Answer
Standard XII
Mathematics
Vector Triple Product
Find the vect...
Question
Find the vector equations of the following planes in scalar product form
r
→
·
n
→
=
d
:
(i)
r
→
=
2
i
^
-
k
^
+
λ
i
^
+
μ
i
^
-
2
j
^
-
k
^
(ii)
r
→
=
1
+
s
-
t
t
^
+
2
-
s
j
^
+
3
-
2
s
+
2
t
k
^
(iii)
r
→
=
i
^
+
j
^
+
λ
i
^
+
2
j
^
-
k
^
+
μ
-
i
^
+
j
^
-
2
k
^
(iv)
r
→
=
i
^
-
j
^
+
λ
i
^
+
j
^
+
k
^
+
μ
4
i
^
-
2
j
^
+
3
k
^
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Solution
i
We know that the equation
r
→
=
a
→
+
λ
b
→
+
μ
c
→
represents a plane passing through a point whose position vector is
a
→
and parallel to the vectors
b
→
and
c
→
.
Here,
a
→
=
2
i
^
+
0
j
^
-
k
^
;
b
→
=
i
^
;
c
→
=
i
^
-
2
j
^
-
k
^
Normal vector,
n
→
=
b
→
×
c
→
=
i
^
j
^
k
^
1
0
0
1
-
2
-
1
=
0
i
^
+
j
^
-
2
k
^
=
j
^
-
2
k
^
The vector equation of the plane in scalar product form is
r
→
.
n
→
=
a
→
.
n
→
⇒
r
→
.
j
^
-
2
k
^
=
2
i
^
+
0
j
^
-
k
^
.
j
^
-
2
k
^
⇒
r
→
.
j
^
-
2
k
^
=
2
i
i
The given equation of the plane is
r
→
=
1
+
s
-
t
i
^
+
2
-
s
j
^
+
3
-
2
s
+
2
t
k
^
⇒
r
→
=
i
^
+
2
j
^
+
3
k
^
+
s
i
^
-
j
-
2
k
^
+
t
-
i
^
+
0
j
^
+
2
k
^
We know that the equation
r
→
=
a
→
+
s
b
→
+
t
c
→
represents a plane passing through a point whose position vector is
a
→
and parallel to the vectors
b
→
and
c
→
.
Here,
a
→
=
i
^
+
2
j
^
+
3
k
^
;
b
→
=
i
^
-
j
-
2
k
^
;
c
→
=
-
i
^
+
0
j
^
+
2
k
^
Normal vector,
n
→
=
b
→
×
c
→
=
i
^
j
^
k
^
1
-
1
-
2
-
1
0
2
=
-
2
i
^
+
0
j
^
-
k
^
=
-
2
i
^
-
k
^
The vector equation of the plane in scalar product form is
r
→
.
n
→
=
a
→
.
n
→
⇒
r
→
.
-
2
i
^
-
k
^
=
i
^
+
2
j
^
+
3
k
^
.
-
2
i
^
-
k
^
⇒
r
→
.
-
1
2
i
^
+
k
^
=
-
2
+
0
-
3
⇒
r
→
.
-
1
2
i
^
+
k
^
=
-
5
⇒
r
→
.
2
i
^
+
k
^
=
5
i
i
i
We know that the equation
r
→
=
a
→
+
λ
b
→
+
μ
c
→
represents a plane passing through a point whose position vector is
a
→
and parallel to the vectors
b
→
and
c
→
.
Here,
a
→
=
i
^
+
j
^
+
0
k
^
;
b
→
=
i
^
+
2
j
^
-
k
^
;
c
→
=
-
i
^
+
j
^
-
2
k
^
Normal vector,
n
→
=
b
→
×
c
→
=
i
^
j
^
k
^
1
2
-
1
-
1
1
-
2
=
-
3
i
^
+
3
j
^
+
3
k
^
The vector equation of the plane in scalar product form is
r
→
.
n
→
=
a
→
.
n
→
⇒
r
→
.
-
3
i
^
+
3
j
^
+
3
k
^
=
i
^
+
j
^
+
0
k
^
.
-
3
i
^
+
3
j
^
+
3
k
^
⇒
r
→
.
-
3
i
^
+
3
j
^
+
3
k
^
=
-
3
+
3
⇒
r
→
.
3
-
i
^
+
j
^
+
k
^
=
0
⇒
r
→
.
-
i
^
+
j
^
+
k
^
=
0
i
v
We know that the equation
r
→
=
a
→
+
λ
b
→
+
μ
c
→
represents a plane passing through a point whose position vector is
a
→
and parallel to the vectors
b
→
and
c
→
.
Here,
a
→
=
i
^
-
j
^
+
0
k
^
;
b
→
=
i
^
+
j
^
+
k
^
;
c
→
=
4
i
⏜
-
2
j
^
+
3
k
^
Normal vector,
n
→
=
b
→
×
c
→
=
i
^
j
^
k
^
1
1
1
4
-
2
3
=
5
i
^
+
j
^
-
6
k
^
The vector equation of the plane in scalar product form is
r
→
.
n
→
=
a
→
.
n
→
⇒
r
→
.
5
i
^
+
j
^
-
6
k
^
=
i
^
-
j
^
+
0
k
^
.
5
i
^
+
j
^
-
6
k
^
⇒
r
→
.
5
i
^
+
j
^
-
6
k
^
=
5
-
1
+
0
⇒
r
→
.
5
i
^
+
j
^
-
6
k
^
=
4
⇒
r
→
.
5
i
^
+
j
^
-
6
k
^
=
4
Disclaimer: The answer given for part (iv) of this problem in the text book is incorrect.
Suggest Corrections
0
Similar questions
Q.
Find the shortest distance between the following pairs of parallel lines whose equations are:
(i)
r
→
=
i
^
+
2
j
^
+
3
k
^
+
λ
i
^
-
j
^
+
k
^
and
r
→
=
2
i
^
-
j
^
-
k
^
+
μ
-
i
^
+
j
^
-
k
^
(ii)
r
→
=
i
^
+
j
^
+
λ
2
i
^
-
j
^
+
k
^
and
r
→
=
2
i
^
+
j
^
-
k
^
+
μ
4
i
^
-
2
j
^
+
2
k
^
Q.
Find the shortest distance between the following pairs of lines whose vector equations are:
(i)
r
→
=
3
i
^
+
8
j
^
+
3
k
^
+
λ
3
i
^
-
j
^
+
k
^
and
r
→
=
-
3
i
^
-
7
j
^
+
6
k
^
+
μ
-
3
i
^
+
2
j
^
+
4
k
^
(ii)
r
→
=
3
i
^
+
5
j
^
+
7
k
^
+
λ
i
^
-
2
j
^
+
7
k
^
and
r
→
=
-
i
^
-
j
^
-
k
^
+
μ
7
i
^
-
6
j
^
+
k
^
(iii)
r
→
=
i
^
+
2
j
^
+
3
k
^
+
λ
2
i
^
+
3
j
^
+
4
k
^
and
r
→
=
2
i
^
+
4
j
^
+
5
k
^
+
μ
3
i
^
+
4
j
^
+
5
k
^
(iv)
r
→
=
1
-
t
i
^
+
t
-
2
j
^
+
3
-
t
k
^
and
r
→
=
s
+
1
i
^
+
2
s
-
1
j
^
-
2
s
+
1
k
^
(v)
r
→
=
λ
-
1
i
^
+
λ
+
1
j
^
-
1
+
λ
k
^
and
r
→
=
1
-
μ
i
^
+
2
μ
-
1
j
^
+
μ
+
2
k
^
(vi)
r
→
=
2
i
^
-
j
^
-
k
^
+
λ
2
i
^
-
5
j
^
+
2
k
^
and
,
r
→
=
i
^
+
2
j
^
+
k
^
+
μ
i
^
-
j
^
+
k
^
(vii)
r
→
=
i
^
+
j
^
+
λ
2
i
^
-
j
^
+
k
^
and
,
r
→
=
2
i
^
+
j
^
-
k
^
+
μ
3
i
^
-
5
j
^
+
2
k
^
(viii)
r
→
=
8
+
3
λ
i
^
-
9
+
16
λ
j
^
+
10
+
7
λ
k
^
and
r
→
=
15
i
^
+
29
j
^
+
5
k
^
+
μ
3
i
^
+
8
j
^
-
5
k
^
[NCERT EXEMPLAR]
Q.
The equation of the plane
r
→
=
i
^
-
j
^
+
λ
i
^
+
j
^
+
k
^
+
μ
i
^
-
2
j
^
+
3
k
^
in scalar product form is
(a)
r
→
·
5
i
^
-
2
j
^
-
3
k
^
=
7
(b)
r
→
·
5
i
^
+
2
j
^
-
3
k
^
=
7
(c)
r
→
·
5
i
^
-
2
j
^
+
3
k
^
=
7
(d) None of these
Q.
Find the shortest distance between the lines
(i)
r
→
=
i
^
+
2
j
^
+
k
^
+
λ
i
^
-
j
^
+
k
^
and
,
r
→
=
2
i
^
-
j
^
-
k
^
+
μ
2
i
^
+
j
^
+
2
k
^
(ii)
x
+
1
7
=
y
+
1
-
6
=
z
+
1
1
and
x
-
3
1
=
y
-
5
-
2
=
z
-
7
1
(iii)
r
→
=
i
^
+
2
j
^
+
3
k
^
+
λ
i
^
-
3
j
^
+
2
k
^
and
r
→
=
4
i
^
+
5
j
^
+
6
k
^
+
μ
2
i
^
+
3
j
^
+
k
^
(iv)
r
→
=
6
i
^
+
2
j
^
+
2
k
^
+
λ
i
^
-
2
j
^
+
2
k
^
and
r
→
=
-
4
i
^
-
k
^
+
μ
3
i
^
-
2
j
^
-
2
k
^
Q.
Find the value of λ so that the following vectors are coplanar:
(i)
a
→
=
i
^
-
j
^
+
k
^
,
b
→
=
2
i
^
+
j
^
-
k
^
,
c
→
=
λ
i
^
-
j
^
+
λ
k
^
(ii)
a
→
=
2
i
^
-
j
^
+
k
^
,
b
→
=
i
^
+
2
j
^
-
3
k
^
,
c
→
=
λ
i
^
+
λ
j
^
+
5
k
^
(iii)
a
→
=
i
^
+
2
j
^
-
3
k
^
,
b
→
=
3
i
^
+
λ
j
^
+
k
^
,
c
→
=
i
^
+
2
j
^
+
2
k
^
(iv)
a
→
=
i
^
+
3
j
^
,
b
→
=
5
k
^
,
c
→
=
λ
i
^
-
j
^
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