3-Dimensional Geometry

Vishnu Shrimal Sir

eSaral Mathematics Faculty
➢ IIT-BHU, Varanasi, (B.Tech)
➢ Ex-Faculty, Allen Career Institute
➢ 7+ years of Teaching Experience
➢ Mentored over thousands Students
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Sonish Singhal
First eSaralite
 MATHS Total MATHS Total MATHS Total
Sequence & Series 12 P&C 4 PMI 0
AOD 11 MOD (Diff.) 4 Relation 0
Binomial Theorm 8 Indef. Integration 4 Logarithm 0
Def. Integration 7 Trigo Ratios &
3
Vectors 7 Identities
Complex Number 6 Point 3
MR 6 Limit 3
Differential Eqn. 6 Hyperbola 2
Determinant 6 Sets 2
Continuity 2
3-Dimensional 6 Trigo Equation 1
Probability 6 Straight Line 1
Statistics 5 ITF 1
Function 5 Differentiability 1
Area Under Curve 5 Solution of
0
Matrices 5 Triangles
Quadratic Equation 4 Height and
0
Circle 4 Distances
Parabola 4
Ellipse 4
Q) The distance of the point (1, –5, 9) from the plane x – y + z = 5
measured along the line x = y = z is : [JEE(Main)-2016]
Sol. Equ a tionof line pa ra llelto x= y = z throu g h 𝟐𝟎
𝐱−𝟏 𝐲+𝟓 𝐳−𝟗 (1)
(1,–5,9) is = = = 𝛌 𝟑
𝟏 𝟏 𝟏
(2) 𝟑 𝟏𝟎
If P( + 1,  – 5,  + 9) be point of
intersection of line and plane. (1, –5, 9) (3) 𝟏𝟎 𝟑

+-+5++9=5 P 𝟏𝟎
(4)
  = -0 𝟑
 Coordinates point are (–9, –15, –1)
 Required distance = 𝟏𝟎 𝟑
 𝐱−𝟑 𝐲+𝟐 𝐳+𝟒
Q) If the line, = = lies in the plane,
𝟐 −𝟏 𝟑
lx + my – z = 9, then l2 + m2 is equal to :− [JEE(Main)-2016]
𝒙−𝟑 𝒚+𝟐 z+𝟒 (1) 2
Sol. Given line = =
𝟐 −𝟏 𝟑
and Given plane is lx + my – z = 9 (2) 26
Now, it is given that line lies on plane (3) 18
\ 2l – m – 3 = 0  2l – m = 3 ...(1)
(4) 5
Also, (3, –2, –4) lies on plane
3l – 2m = 5 ...(2)
Solving (1) and (2), we get
l = 1, m = –1
\ l2 + m2 = 2
Q) The distantce of the point (1, 3, –7) from the plane passing through the point
𝐱– 𝟏 𝐲+𝟐
(1, –1, –1), having normal perpendicular to both the lines =
𝟏 −𝟐
𝐳−𝟒 𝐱– 𝟐 𝐲+𝟏 𝐳+𝟕
= and = = , is :− [JEE(Main)-2017]
𝟑 𝟐 −𝟏 –𝟏
Sol. 𝐢Ƹ 𝐣Ƹ 𝐤መ
Normal vector 𝟏 መ 𝟏𝟎 𝟐𝟎
−𝟐 𝟑 = 𝟓𝐢Ƹ + 𝟕𝐣Ƹ + 𝟑𝐤 (1) (2)
𝟐 −𝟏 −𝟏 𝟕𝟒 𝟕𝟒

So plane is 5(x – 1) + 7(y + 1) + 3(z + 1) = 0 𝟏𝟎 𝟓

(3) (4)
 5x + 7y + 3z + 5 = 0 𝟖𝟑 𝟖𝟑
𝟓 + 𝟐𝟏 − 𝟐𝟏 + 𝟓 𝟏𝟎
𝐃𝐢𝐬𝐭𝐚𝐧𝐜𝐞 =
𝟐𝟓 + 𝟒𝟗 + 𝟗 𝟖𝟑
Q) The length of the projection of the line segment joining the points
(5, –1, 4) and (4, –1, 3) on the plane, x + y + z = 7 is :
[JEE(Main)-2018]
෢
Sol. AC = AB ⋅ AC 𝟐 𝟏
C B(5, –1, 4)
መ (1) (2)
𝐢Ƹ + 𝐣Ƹ + 𝐤 𝟐 𝟑 𝟑
መ ⋅
= (𝐢Ƹ + 𝐤) =
A(4, –1, 3)
𝟑 𝟑
A B
𝟐 𝟐
(3) (4)
𝟑 𝟑
𝟒 𝟐
𝐀′ 𝐁′ = 𝐁𝐂 = 𝐀𝐁𝟐 − 𝐀𝐂 𝟐 = 𝟐− =
𝟑 𝟑

𝟐
Leng th of projection=
𝟑
Q) The plane through the intersection of the planes x + y + z = 1 and
2x + 3y – z + 4 = 0 and parallel to y-axis also passes through the point :
[JEE (MAIN)-2019]
Sol. (𝐱 + 𝐲 + 𝐳 − 𝟏) + 𝛌(𝟐𝐱 + 𝟑𝐲 − 𝐳 + 𝟒) = 𝟎
⇒ (𝟏 + 𝟐𝛌)𝐱 + (𝟏 + 𝟑𝛌)𝐲 + (𝟏 − 𝛌)𝐳 − 𝟏 + 𝟒𝛌 = 𝟎 (A) (-3,0,-1)
dr's of normal of the plane are (B) (3,3,-1)
𝟏 + 𝟐𝛌, 𝟏 + 𝟑𝛌, 𝟏 − 𝛌
(C) (3,3,-1)
Since plane is parallel to y-axis, 𝟏 + 𝟑𝛌 = 𝟎
⇒ 𝛌 = −𝟏 𝟑 Τ (D) (-3,1,1)
So the equation of plane is
𝐱 + 𝟒𝐳 − 𝟕 = 𝟎
Point (3,2,1) satisfies this equation
Q) If the lines x = ay + b, z = cy + d and x = a’z + b’, y = c’zz + d’ are
perpendicular, then [JEE (MAIN)-2019]
Sol. Line 𝐱 = 𝐚𝐲 + 𝐛, 𝐳 = 𝐜𝐲 + 𝐝
𝐱−𝐛 𝐲 𝐳−𝐝 (A) cc’ + a + a’ = 0
⇒ = =
𝐚 𝟏 𝐜
Line 𝐱 = 𝐚′ 𝐳 + 𝐛′ , 𝐲 = 𝐜 ′ 𝐳 + 𝐝′ (B) aa’ + c + c’ = 0
x − b′ y − d′ z
⇒ ′
= ′
= (C) ab’ + bc’ + 1 = 0
a c 𝟏
Given both the lines are perpendicular
(D) bb’ + cc’ + 1 = 0
⇒ aa′ + c ′ + c = 𝟎
Q) Find image of point (2,1,6) in the plane containing points (2,1,0),(6,3,3)
and (5,2,2)
[JEE (MAIN)-2020]
Sol.
Plane is 𝐱 + 𝐲 − 𝟐𝐳 = 𝟑 (1) (6,5,-2)
𝐱−𝟐 𝐲−𝟏 𝐳−𝟔
⇒ = = (2) (6,-5,2)
𝟏 𝟏 −𝟐
−𝟐(𝟐 + 𝟏 − 𝟏𝟐 − 𝟑)
= (3) (2,-3,4)
𝟔
⇒ (𝐱, 𝐲, 𝐳) = (𝟔, 𝟓, −𝟐) (4) (2,-5,6)
Q) Shortest distance between the lines
𝐱−𝟑 𝐲−𝟖 𝐳−𝟑 𝐱+𝟑 𝐲+𝟕 𝐳−𝟔
= = , = = 𝐢𝐬 − [JEE (MAIN)-2020]
𝟏 𝟒 𝟐𝟐 𝟏 𝟏 𝟕
Sol.
AB = 𝟔𝐢Ƹ + 𝟏𝟓𝐣Ƹ + 𝟑𝐤መ (1) 𝟑 𝟑𝟎
p = መi + 𝟒መj + 𝟐𝟐k෠ (2) 𝟐 𝟑𝟎
q = መi + መj + 𝟕k෠
(3) 𝟑𝟎
𝐢 𝐣 𝐤
𝐩 × 𝐪 = 𝟏 𝟒 𝟐𝟐 = 𝟔𝐢Ƹ + 𝟏𝟓𝐣Ƹ − 𝟑𝐤 መ
(4) 𝟒 𝟑𝟎
𝟏 𝟏 𝟕
|𝐀𝐁 ⋅ (𝐩 × 𝐪൯ |𝟑𝟔 + 𝟐𝟐𝟓 + 𝟗|
𝐒. 𝐃. = = = 𝟑 𝟑𝟎
|𝐩 × 𝐪| 𝟑𝟔 + 𝟐𝟐𝟓 + 𝟗
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