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Vector Algebra
NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Vector Algebra
NDA 2024 · Mathematics · Vector Algebra
What is 3 α+2 β equal to if (2 i +6 j +27 k ) ×( i +α j +β k ) is a...
What is
3
α
+
2
β
3 \alpha+2 \beta
3
α
+
2
β
equal to if
(
2
i
^
+
6
j
^
+
27
k
^
)
×
(
i
^
+
α
j
^
+
β
k
^
)
(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\alpha \hat{j}+\beta \hat{k})
(
2
i
^
+
6
j
^
+
27
k
^
)
×
(
i
^
+
α
j
^
+
β
k
^
)
is a null vector?
Options
A
36
36
36
B
33
33
33
C
30
30
30
D
27
27
27
Correct answer
A.
36
36
36
Step-by-step solution
Given that,
(
2
i
^
+
6
j
^
+
27
k
^
)
×
(
i
^
+
α
j
^
+
β
k
^
)
=
0
(2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+27 \hat{\mathrm{k}}) \times(\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\beta \hat{\mathrm{k}})=0
(
2
i
^
+
6
j
^
+
27
k
^
)
×
(
i
^
+
α
j
^
+
β
k
^
)
=
0
⇒
∣
i
^
j
^
k
^
2
6
27
1
α
β
∣
=
0
⇒
i
^
(
6
β
−
27
α
)
−
j
^
(
2
β
−
27
)
+
k
^
(
2
α
−
6
)
=
0
\begin{aligned} & \Rightarrow\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \alpha & \beta \end{array}\right|=0 \\ & \Rightarrow \hat{i}(6 \beta-27 \alpha)-\hat{j}(2 \beta-27)+\hat{k}(2 \alpha-6)=0 \end{aligned}
⇒
i
^
2
1
j
^
6
α
k
^
27
β
=
0
⇒
i
^
(
6
β
−
27
α
)
−
j
^
(
2
β
−
27
)
+
k
^
(
2
α
−
6
)
=
0
C
o
m
p
a
r
i
n
g
b
o
t
h
s
i
d
e
s
o
f
a
b
o
v
e
e
q
u
a
t
i
o
n
,
w
e
g
e
t
:
−
Comparing both sides of above equation, we get:-
C
o
m
p
a
r
in
g
b
o
t
h
s
i
d
eso
f
ab
o
v
ee
q
u
a
t
i
o
n
,
w
e
g
e
t
:
−
6
β
−
27
α
=
0
⇒
2
β
=
9
α
2
β
−
27
=
0
⇒
β
=
27
2
and
2
α
−
6
=
0
⇒
α
=
3
\begin{aligned} & 6 \beta-27 \alpha=0 \Rightarrow 2 \beta=9 \alpha \\ & 2 \beta-27=0 \Rightarrow \beta=\frac{27}{2} \text{ and } 2 \alpha-6=0 \Rightarrow \alpha=3 \end{aligned}
6
β
−
27
α
=
0
⇒
2
β
=
9
α
2
β
−
27
=
0
⇒
β
=
2
27
and
2
α
−
6
=
0
⇒
α
=
3
S
o
,
So,
S
o
,
3 \alpha+2 \beta=3 \times 3+2 \times \frac{27}{2}=9+27=36$
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