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TS EAMCET 19-July-2022 Shift 1 Solved Paper
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© examsnet.com
Question : 30
Total: 160
If
(
α
,
β
,
γ
)
is a triad of real numbers satisfying
i
−
2
j
+
5
k
=
α
(
i
+
j
+
k
)
+
β
(
i
+
2
j
+
3
k
)
+
γ
(
2
i
−
j
+
k
)
, then
α
2
−
β
2
+
γ
2
=
23
31
40
-6
Validate
Solution:
👈: Video Solution
^
i
−
2
^
j
+
5
^
k
=
α
(
^
i
+
^
j
+
^
k
)
+
β
(
^
i
+
2
^
j
+
3
^
k
)
+
γ
(
2
^
i
−
^
j
+
^
k
)
⇒
1
=
α
+
β
+
2
γ
......(i)
−
2
=
α
+
2
β
−
γ
......(ii)
5
=
α
+
3
β
+
γ
.......(iii)
From Eqs. (i) and (ii), we get
−
β
+
3
γ
=
3
.....(iv)
From Eqs. (ii) and (iii), we get
β
+
2
γ
=
7
......(v)
From Eqs. (iv) and (v), we get
γ
=
2
,
and
β
=
3
α
=
−
6
α
2
−
β
2
+
γ
2
=
36
−
9
+
4
=
31
© examsnet.com
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