A unit vector perpendicular to vectors i-2j+k and 3i+j-2k is
(Or)
A unit vector perpendicular to i^-2j^+k^ and 3i^+j^-2k^
Answers
Answered by
1
Answer:
Explanation:A vector perpendicular to
2
vectors is calculated with the determinant (cross product of 2 vectors)
∣
∣
∣
∣
∣
→
i
→
j
→
k
d
e
f
g
h
i
∣
∣
∣
∣
∣
where
→
a
=
⟨
d
,
e
,
f
⟩
and
→
b
=
⟨
g
,
h
,
i
⟩
are the 2 vectors
Here, we have
→
a
=
⟨
1
,
−
2
,
2
⟩
and
→
b
=
⟨
3
,
1
,
−
2
⟩
Therefore,
∣
∣
∣
∣
∣
→
i
→
j
→
k
1
−
2
2
3
1
−
2
∣
∣
∣
∣
∣
=
→
i
∣
∣
∣
−
2
2
1
−
2
∣
∣
∣
−
→
j
∣
∣
∣
1
2
3
−
2
∣
∣
∣
+
→
k
∣
∣
∣
1
−
2
3
1
∣
∣
∣
=
→
i
(
(
−
2
)
⋅
(
−
2
)
−
(
2
)
⋅
(
1
)
)
−
→
j
(
(
1
)
⋅
(
−
2
)
−
(
2
)
⋅
(
3
)
)
+
→
k
(
(
1
)
⋅
(
1
)
−
(
−
2
)
⋅
(
3
)
)
=
⟨
2
,
8
,
7
⟩
=
→
c
Verification by doing 2 dot products
⟨
1
,
−
2
,
2
⟩
.
⟨
2
,
8
,
7
⟩
=
(
1
)
⋅
(
2
)
+
(
−
2
)
⋅
(
8
)
+
(
2
)
⋅
(
7
)
=
0
⟨
3
,
1
,
−
2
⟩
.
⟨
2
,
8
,
7
⟩
=
(
3
)
⋅
(
2
)
+
(
1
)
⋅
(
8
)
+
(
−
2
)
⋅
(
7
)
=
0
So,
→
c
is perpendicular to
→
a
and
→
b
The unit vector is
ˆ
c
=
→
c
∣
∣
∣
∣
→
c
∣
∣
∣
∣
∣
∣
∣
∣
→
c
∣
∣
∣
∣
=
|
|
<
2
,
8
,
7
>
|
|
=
√
2
2
+
8
2
+
7
2
=
√
117
The unit vector is
ˆ
c
=
1
√
117
<
2
,
8
,
7
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