TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Let the unit vector be
‌a=x

∧

i

+y

∧

j

+z

∧

k

. . . (i)
Then, ‌‌x2+y2+z2=1‌‌‌‌... (ii)
‌x2+y2+z2=1
‌b=

∧

i

−2

∧

j

+3

∧

k

‌c=

∧

i

+

∧

j

+

∧

k

‌d=2

∧

i

−

∧

j

−

∧

k

‌a⟂b⇒a⋅b=0
‌x−2y+3z=0 . . . (iii)
a, c and d are coplanar vectors.
∴‌‌|

x	y	z
1	1	1
2	−1	−1

|=0
⇒‌‌x(−1+1)−y(−1−2)+z(−1−2)=0
⇒‌‌3y−3z=0⇒y=z
From Eq. (iii),
x−2z+3z‌‌=0
x+z‌‌=0
x‌‌=−z
⇒ or From Eq. (ii),
‌x2+x2+x2=1
‌x=±‌

1

√3

⇒y=∓‌

1

√3

‌ and ‌z=∓‌

1

√3

∴ Required unit vector =±‌

1

√3

(

∧

i

−

∧

j

−

∧

k

)