Find unit vector perpendicular to the plane of a=2iˆ−2jˆ+kˆ and b=iˆ+2jˆ+2kˆ
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Answer:
We know that cross product of any two vectors yields a vector which is perpendicular to both vectors
∴
for two vectors
→
A
and
→
B
if
→
C
is the vector perpendicular to both.
→
C
=
→
A
×
→
B
=
⎡
⎢
⎣
ˆ
i
ˆ
j
ˆ
k
A
1
A
2
A
3
B
1
B
2
B
3
⎤
⎥
⎦
=
(
A
2
B
3
−
B
2
A
3
)
ˆ
i
−
(
A
1
B
3
−
B
1
A
3
)
ˆ
j
+
(
A
1
B
2
−
B
1
A
2
)
ˆ
k
.
Inserting given vectors we obtain
→
C
=
⎡
⎢
⎣
ˆ
i
ˆ
j
ˆ
k
2
1
1
1
−
1
2
⎤
⎥
⎦
=
(
1
×
2
−
(
−
1
)
×
1
)
ˆ
i
−
(
2
×
2
−
1
×
1
)
ˆ
j
+
(
2
×
(
−
1
)
−
1
×
1
)
ˆ
k
.
=
3
ˆ
i
−
3
ˆ
j
−
3
ˆ
k
.
Now unit vector in the direction of
→
C
is
→
C
∣
∣
∣
→
C
∣
∣
∣
∴
∣
∣
∣
→
C
∣
∣
∣
=
√
3
2
+
(
−
3
)
2
+
(
−
3
)
2
=
√
27
=
3
√
3
Therefore desired unit vector is
1
√
3
(
ˆ
i
−
ˆ
j
−
ˆ
k
)
Explanation:
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