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NDA Math Vector Algebra
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© examsnet.com
Question : 54
Total: 101
For any vector
→
a
|
→
a
×
⋀
x
|
2
+
|
→
a
×
⋀
j
|
2
+
|
→
a
×
⋀
k
|
2
is equal to
[NDA II 2017]
|
→
a
|
2
2
|
→
a
|
2
3
|
→
a
|
2
4
|
→
a
|
2
Validate
Solution:
Let
→
a
=
x
⋀
i
+
y
⋀
j
+
3
⋀
k
|
→
a
×
⋀
i
|
2
=
z
2
+
y
2
|
→
a
×
⋀
j
|
2
=
x
2
+
z
2
|
→
a
×
⋀
k
|
2
=
x
2
+
y
2
∴
2
(
x
2
+
y
2
+
z
2
)
=
2
|
→
a
|
2
© examsnet.com
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