Vector 3D
JEE Main 2021 (February)
Hints and Solutions
MathonGo
Q1 (3)
∣ ^i
∣
Normal vector of required plane is n→ = ∣ 3
∣
∣2
^j
1
−5
k^ ∣
∣
−2 ∣ = −11^i − ^j − 17k^
∣
−1 ∣
∴ 11(x − 1) + (y − 2) + 17(z + 3) = 0
11x + y + 17z + 38 = 0
Q2 (1)
x−3
1
=
y−4
2
=
z−5
2
=λ
⇒ x = λ + 3, y = 2λ + 4, z = 2λ + 5
Which lines on given plane hence
⇒ λ + 3 + 2λ + 4 + 2λ + 5 = 17
⇒λ=
5
5
=1
Hence, point of intersection is Q(4, 6, 7)
∴ Required distance = PQ
= √9 + 25 + 4
= √38
Q3 (2)
P (a, 6, 9), Q(20, b, −a − 9)
mid point of PQ = ( a+20
,
2
b+6
2
, − 2a )
lie on line
a+20
2
−3
=
b+6
2
−2
7
5
a+20−6
= b+6−4
14
10
a+14
= a+2
14
18
=
=
− 2a −1
−9
−a−2
−18
18a + 252 = 14a + 28
#PaperPhodnaHai
www.mathongo.com
Vector 3D
JEE Main 2021 (February)
Hints and Solutions
MathonGo
4a = −224
a = −56
b+2
10
b+2
10
b+2
10
a+2
18
−54
18
=
=
= −3 ⇒ b = −32
|a + b| = | − 56 − 32| = 88
Q4 (2)
^ − 1} + λ{r→ ⋅ (^i − 2^j) + 2} = 0
Plane passing through intersection of plane is {r→ ⋅ (^i + ^j + k)
^ we get
Passes through ^i + 2k,
(3 − 1) + λ(1 + 2) = 0 ⇒ λ = − 23
^ − 1} − 2{r→ ⋅ (^i − 2^j) + 2} = 0
Hence, equation of plane is 3{r→ ⋅ (^i + ^j + k)
→
^ =7
⇒ r ⋅ (^i + 7^j + 3k)
Q5 (1)
1
x−λ
1
=
x−λ
2
=
x
1
=
y− 2
1
2
=
z
1
−2
=
2
−1
1
y− 2
1
y+2λ
1
=
z−λ
1
… . . (1)
1
2
=
+ 2λ
∣∣−5λ− 3 ∣∣
2
√14
→ → → →
[ a 2− a 1 b 1 b 2]
∣→ → ∣
∣ b 1× b 2∣
∣
∣
−λ ∣
∣
−1 ∣
∣
1 ∣
1
1
^j
1
1
2
Point on line = (0, −2λ, λ)
Distance between skew lines =
∣λ
∣
∣2
∣
∣1
∣ ^i
∣
∣2
∣
∣1
Point on line = (λ, 1 , 0)
k^ ∣
∣
−1 ∣
∣
1 ∣
=
√7
2√ 2
(given)
= |10λ + 3| = 7 ⇒ λ = −1
⇒ |λ| = 1
Q6 (3)
#PaperPhodnaHai
www.mathongo.com
Vector 3D
JEE Main 2021 (February)
Hints and Solutions
MathonGo
| 2 + m2 + n 2 = 1
1
√2
∴ 2n2 = 1 ⇒ n = ±
∴|2 + m2 =
⇒
1
2
1
&
2
∣ +m =
1
√2
1
2
− 2 lm =
⇒ 1 m = 0 or m = 0
1
√2
∴ I = 0, m =
< 0,
1
√2
,
1
√2
>
or I =
or <
∴ cos α = 0 + 0 +
1
2
=
1
√2
1
2
1
√2
, 0,
1
√2
>
∴ sin4 α + cos4 α = 1 − 12 sin2 (2α) = 1 − 12 ,
3
4
=
5
8
Q7 (1)
x−1
2
=
y+1
3
=
z−1
−2
=λ
Any point on this line (2λ + 1, 3λ − 1, −2λ + 1)
Direction ratio of given line (2,3,-2)
→ →
Direction ratio of line to be found (2λ + 1, 3λ − 2, −2λ − 1) ∴ d 1 ⋅ d 2 = 0
λ = 2/17
Direction ratio of line (21,-28,-21)≡ (3, −4, −3) ≡ (−3, 4, 3)
Q8 (3)
A(1, 2, 3), B(2, 3, 1), C(2, 4, 2), O(0, 0, 0)
Equation of plane passing through A, B, C will be
#PaperPhodnaHai
www.mathongo.com
Vector 3D
JEE Main 2021 (February)
Hints and Solutions
MathonGo
∣x−1 y−2 z−3 ∣
∣2−1 3−2 1−3 ∣ = 0
∣
∣
∣2−1 4−2 2−3 ∣
∣ x−1 y−2 z−3 ∣
⇒∣∣ 1
1
−2 ∣∣ = 0
∣ 1
2
−1 ∣
⇒(x − 1)(−1 + 4) − (y − 2)(−1 + 2) + (z − 3)(2 − 1) = 0
⇒(x − 1)(3) − (y − 2)(1) + (z − 3)(1) = 0
⇒3x − 3 − y + 2 + z − 3 = 0
⇒ 3x − y + z − 4 = 0, is the required plane.
Now, given O(0, 0, 0)&P(2, −1, 1)
Plane is 3x − y + z − 4 = 0
O′ &P′ are foot of perpendiculars.
for O′
y−0
x−0
= −1 = z−0
3
1
y
4
x
z
= −1 = 1 = 11
3
O′ ( 12
,
11
⇒
−4
11
=
,
4
)
11
−(0−0+0−4)
9+1+1
for P′
−(3(2)−(−1)+1−4)
9+1+1
x−2
3
=
y+1
−1
=
z−1
1
=
x−2
3
=
y+1
−1
=
z−1
1
= ( −4
)
11
P ′ ( −12
+ 2,
11
⇒
P ′ ( 10
,
11
4
11
−7
11
O′ P ′ = √( 10
−
11
,
− 1,
7
)
11
2
12
)
11
−4
11
+ 1)
+ ( −7
+
11
4
)
11
2
7
+ ( 11
−
4
)
11
2
#PaperPhodnaHai
www.mathongo.com
Vector 3D
JEE Main 2021 (February)
Hints and Solutions
MathonGo
1
√4 + 9 +
11
√
O′ P ′ = 1122
√ √
O′ P ′ = 2×11 11
2
O′ P ′ = √ 11
⇒ O′ P ′ =
⇒
⇒
⇒
9
Q9 (44)
→
ℓ1 : r = (3 + t)^i + (−1 + 2t)^j + (4 + 2t)k^
ℓ1 :
x−3
1
=
x−3
2
=
y+1
2
=
y−3
2
=
z−4
2
⇒
z−2
1
⇒
D. R. of ℓ1 = 1, 2, 2
→
ℓ2 : r = (3 + 2 s)^i + (3 + 2 s)^j + (2 + s)k^
ℓ2 :
D. R. of ℓ2 = 2, 2, 1
D ⋅ R, of ℓ is ⊥ to ℓ1 &ℓ2
∴
D.R. of ℓ|| (ℓ1 × ℓ2 )
∴ Equation of ℓ :
x
2
=
y
−3
⇒
=
⟨−2, 3, −2⟩
z
2
Solving ℓ&ℓ1
(2λ, −3λ, 2λ) = (μ + 3, 2μ − 1, 2μ + μ)
⇒
2λ = μ + 3
−3λ = 2μ − 1
2λ = 2μ + 4
μ = −1
μ = −1
P(2, −3, 2){ intersection point }
Let, Q(2v + 3, 2v + 3, v + 2) be point on ℓ2
Now, PQ = √(2v + 3 − 2)2 + (2v + 3 + 3)2 + (v + 2 − 2)2 = √17
⇒ (2v + 1)2 + (2v + 6)2 + (v)2 = 17
⇒ 9v2 + 28v + 36 + 1 − 17 = 0
⇒ 9v2 + 28v + 20 = 0
⇒ 9v2 + 18v + 10v + 20 = 0
⇒ (9v + 10)(v + 2) = 0
#PaperPhodnaHai
www.mathongo.com
Vector 3D
JEE Main 2021 (February)
Hints and Solutions
MathonGo
⇒ v = −2( rejected ), − 10
( accepted )
9
Q (3 −
20
,3
9
−
20
,2
9
−
10
)
9
( 79 , 79 , 89 )
∴
18(a + b + c)
= 18 ( 79 +
7
9
+ 89 )
= 44
Q10 (1)
P1 = x + 5y + 7z = 3
P2 = x − 3y − z = 5
P3 = x + 5y + 7z = 5/2
⇒ P1 ∥P3
Q11 (4)
Let P(1, 5, 35), Q(7, 5, 5), R(1, λ, 7), S(2λ, 1, 2)
−→
−→
[PQPR PS] = 0
∣ 6
∣ 0
∣
∣ 2λ − 1
∣ 1
∣ 0
∣
∣ 2λ − 1
0
λ−5
−4
0
λ−5
−4
−30 ∣
−28 ∣∣ = 0
−33 ∣
−5 ∣
−28 ∣∣ = 0
−33 ∣
{−33λ + 165 − 112} + 5(λ − 5)(2λ − 1) = 0
53 − 33λ + 5 {2λ2 − 11λ + 5} = 0
16λ2 − 88λ + 78 = 0
λ
5λ2 − 44λ + 39 = 0 <λ12
⇒ λ1 + λ2 = 44/5
Q12 (8)
#PaperPhodnaHai
www.mathongo.com
Vector 3D
JEE Main 2021 (February)
Hints and Solutions
MathonGo
−→ −−→
AB ⊥ P Q
^ ⋅ [+^i + 5^j − 6k]
^ =0
[(4 − λ)^i − 4^j + k]
4 − λ − 20 − 6 = 0
N = −22
Now,
λ
11
2
= −2
λ
⇒ ( 11
) −
⇒
4λ
11
−4
4+8−4=8
Q13 (4)
∣ ^i
∣
Dr's of line ∣ 1
∣
∣0
^j
2
1
k^ ∣
∣
1 ∣ = 3^i − 2^j + k^
∣
2∣
Dr/s : −(3, −2, 1)
Points on the line (-2,4,0) Equation of the line
x+2
3
=
y−4
−2
=
z
1
=λ
#PaperPhodnaHai
www.mathongo.com
Vector 3D
JEE Main 2021 (February)
Hints and Solutions
MathonGo
Dr's of PQ; 3λ − 5, −2λ + 2. λ − 1
Dr's of y lines are (3,-2,1) Since PQ ⊥ line 3(3λ − 5) − 2(−2λ + 2) + 1(λ − 1) = 0
λ=
10
7
P ( 16
, 8,
7 7
10
)
7
21(α + β + γ) = 21 ( 34
) = 102
7
Q14 (1)
Image of (1,3,5) in the plane 4x − 5y + 2z = 8 is (α, β, γ)
⇒
α−1
4
=
β−3
−5
=
γ−5
2
∴ α = 1 + 4 ( 25 ) =
(4(1)−5(3)+2(5)−8)
42 +52 +22
=
2
5
13
5
β = 3 − 5 ( 25 ) = 1 =
γ = 5 + 2 ( 25 ) =
= −2
5
5
29
5
Thus, 5(α + β + γ) = 5 ( 13
+
5
5
5
+
29
)
5
= 47
#PaperPhodnaHai
www.mathongo.com