TS EAMCET 4 May 2018 Shift 2 Solved Paper

The equation for the vectors containing

^

i

and

^

i

+

^

j

is,
[r−

^

i

,

^

i

,

^

i

+

^

j

]=0
(r−

^

i

)[

^

i

×

^

i

+

^

j

]=0
(r−

^

i

)

^

k

=0
z=0
The equation for the vectors containing

^

i

−

^

j

and

^

i

+

^

k

is,
[r−(

^

i

−

^

j

).[(

^

i

−

^

j

)×(

^

i

+

^

k

)]]=0
(r−

^

i

)[

^

i

×

^

i

+

^

i

×

^

k

−

^

j

×

^

i

−

^

j

×

^

k

]=0
[(x−1)

^

i

+(y+1)

^

j

+z

^

k

](−

^

j

+

^

k

−

^

i

)=0
x+y−z=0
Now let a=a1

^

i

+a2

^

j

+a3

^

k

since a is parallel to the above planes, then,a3=0
a1=−a2
a3=0
This implies,
a=a1(

^

i

−

^

j

)
The angle is given by,
cos‌θ=

1×1+(−2)×(−1)

√1+1√1+4+4

=

1+2

√2×3

=

1

√2

θ=

π

4

or 45°