JEE Advanced 2023 Paper 1 Solutions

P:√3x+2y+3z=16
S=α

^

i

+β

^

j

+γ

^

k

:α2+β2+γ2=1
d(α,β,γ)‌ from ‌P=‌

7

2

|

→

u

−

→

v

|=|

→

v

−

→

w

|=|

→

w

−

→

u

|−
V : volume of parallelepiped by vectors

→

u

,

→

v

,

→

w

‌

80

√3

V=?
d(α,β,γ)‌ from ‌β=‌

7

2

‌ (Given) ‌
⇒‌

|√3α+2β+3γ−16|

√3+4+9

=‌

7

2

=‌

|√3α+2β+3γ−16|

4

=‌

7

2

|√3α+2β+3γ−16|=14...(i)
α2+β2+γ2=1.....(ii)
Volume of parallelepiped by vector

→

u

,

→

v

,

→

w

V‌=[

→

u

→

V

→

w

]
‌=

→

u

⋅(V×w)....(iii)
|

→

u

|=|

→

v

|=|

→

w

|=1 (Given)...(iv)
|

→

u

−

→

v

|=|

→

v

−

→

w

|=|

→

w

−

→

u

|‌-(Given) ‌
⇒|

→

u

−

→

v

|2=|

→

v

−

→

w

|2=|

→

w

−

→

u

|2
⇒

u2+v2

(A)

−2

→

u

⋅

→

v

=

v2+w2

(B)

−2

→

v

⋅

→

w

=

w2+u

(C)

2−2

→

w

⋅

→

u

(A) and (B)
⇒u2+v2−2

→

u

⋅

→

v

=v2+w2−2

→

v

⋅

→

w

⇒u2−w2=2

→

u

⋅

→

v

−2

→

v

⋅

→

w

  [∵|

→

u

|=|

→

w

|=1 (Given)]
⇒

→

u

⋅

→

v

=

→

v

⋅

→

w

Hence, by using (B) and (C) also, we will get

→

u

⋅

→

v

=

→

v

⋅

→

w

=

→

w

⋅

→

u

=m (say)...(v)
⇒

→

u

,

→

v

,

→

w

are the vectors of an equilateral triangle (say △ABC )d(O,P)=frac‌16‌sqrt‌3+4+9
‌=‌

16

4

‌=4‌ units ‌

→

OA

‌=

→

u

,

→

OB

=

→

v

,

→

OC

=

→

w

|

→

OA

|‌=|

→

OB

|=|

→

OC

|=1‌ (Given) ‌
‌=‌

16

4

‌=4‌ units ‌

→

OA

‌=

→

u

,

→

OB

=

→

v

,

→

OC

=

→

w

|

→

OA

|‌=|

→

OB

|=|

→

OC

|=1‌ (Given) ‌
In an equilateral triangle, circumcentre, orthrocentre and centroid coincide.
Let D be the circumcentre of △ABC, then
∠ADB=120∘
Given =‌

DA2+DB2−AB2

2(DA)⋅(DB)

...(vi)
OE=OD+DE
=OD+AF
⇒‌‌4=OD+‌

7

2

⇒OD=4−‌

7

2

=‌

1

2

⇒‌‌DA=√OA2−OD2
‌‌=√1−‌

1

4

DA=‌

√3

2

⇒‌‌DA=DB=‌

√3

2

From (vi) and (vii),
−‌

1

2

=‌

‌ 3 4 + 3 4 −AB2

2‌ √3 2 ×‌ √3 2

−‌

1

2

=‌

‌ 3 2 −AB2

3 2

⇒‌‌−‌

1

2

×‌

3

2

=‌

3

2

−AB2
⇒AB2=‌

3

2

+‌

3

4

⇒AB2=‌

9

4

⇒AB−‌

3

2

=|

→

u

−

→

v

|
⇒AB2=‌

9

4

=u2+v2−2

→

u

−

→

v

⇒‌‌‌

9

4

=1+1−2m
⇒2m=2−‌

9

4

=−‌

1

4

⇒m=−‌

1

8

....(viii)
Volume of parallelepiped,
V=|[

→ u

→ v

→ w

]|
|

→

u

→

v

→

w

|2‌=|

1	→ u ⋅ → v	→ u ⋅ → w
→ u ⋅ → v	1	→ v ⋅ → w
→ w ⋅ → u	→ w ⋅ → v	1

|
‌=|

1	m	m
m	1	m
m	m	1

|
=1(1−m2)−m(m−m2)+m(m2−m)
=1−m2−m2+m3+m3−m2
=1−3m2+2m3
|

→

u

→

v

→

w

|2‌=2m3−3m2+1
‌=(m−1)[2m2−m−1]
=(m−1)[2m2−2m+m−1]
=(m−1)(m−1)(2m+1)
=(m−1)2(2m+1)
⇒|[

→ u

→ v

→ w

]|=(m−1)√(2m+1)=v
|(−‌

1

8

−1)√2×−‌

1

8

+1|
V=‌

9

8

×‌

√3

2

‌

80

√3

v‌=‌

80

√3

×‌

9

8

×‌

√3

2

=‌45