AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

a=

^

i

−

^

j

+

^

k

b=

^

i

−2

^

j

+

^

k

c=p

^

i

+2

^

j

+q

^

k

d=p

^

i

+q

^

j

+2

^

k

As the projection of mathbf‌c on a is 5√3 so,

a.c

|a|

=5√3

p+q−2

√3

=5√3
p+q−2=15
p+q=17 .......(I)
It is also given that,
[

a

b

c

]=5
So,
[

1	−1	1
1	−2	1
p	2	q

]=5
1(−2q−2)+1(q−p)+1(2+2p)=5
−2q−2+q−p+2+2p=5
p−q=5 ......(II)
From equation (I) and (II),
p=11,q=6
Now,
b.d=(

^

i

−2

^

j

+

^

k

).(p

^

i

+q

^

j

+2

^

k

)
=p−2q+2
=11−12+2
=1
This gives,
tan−1(b.d)=tan−1(1)
=

π

4