3) Prove that
a= [ 1/0 ] is
is a unit vector
Answers
Answer:Let
a
=
7
1
(2
i
^
+3
j
^
+6
k
^
)=
7
2
i
^
+
7
3
j
^
+
7
6
k
^
,
b
=
7
1
(3
i
^
−6
j
^
+2
k
^
)=
7
3
i
^
−
7
6
j
^
+
7
2
k
^
,
and
c
=
7
1
(6
i
^
+2
j
^
−3
k
^
)=
7
6
i
^
+
7
2
j
^
−
7
3
k
^
.
∣
a
∣=
(
7
2
)
2
+(
7
3
)
2
+(
7
6
)
2
=
49
4
+
49
9
+
49
36
=1
∣
b
∣=
(
7
3
)
2
+(−
7
6
)
2
+(
7
2
)
2
=
49
9
+
49
36
+
49
9
=1
∣
c
∣=
(
7
6
)
2
+(
7
2
)
2
+(−
7
3
)
2
=
49
36
+
49
4
+
49
9
=1
Thus, each of the given three vectors is a unit vector.
a
⋅
b
=
7
2
×
7
3
+
7
3
×(
7
−6
)+
7
6
×
7
2
=
49
6
−
49
18
+
49
12
=0
b
⋅
c
=
7
3
×
7
6
+(
7
−6
)×
7
2
+
7
2
×(
7
−3
)=
49
18
−
49
12
−
49
6
=0
c
⋅
a
=
7
6
×
7
2
+
7
2
×
7
3
+(
7
−3
)×
7
6
=
49
12
+
49
6
−
49
18
=0
Hence, the given three vectors are mutually perpendicular to each other.
Step-by-step explanation: