JEE Advanced Level Test: Vector - JEE MCQ

a = 2(2)^½ i -  j + 4k
|b| = 10
[(2(2λ)² + (1λ)² + (4λ)²]^½
=[ 8λ² + λ² + 16λ²]^½ = 10
= 25λ² = (10)²
25λ² = 100
λ = +-2
b = +-(4(2)^½ i - 2j + 8k)
Therefore, 2a +- b

Let AD is the bisector of ∠A.Then,
BD/DC = AB/ACeq(1)
Given the vertices of triangle,
AB = √3² + 2² + 1² = √14
AC = √1² + 2² + 3² = √14
As AB = BC, So, BD = DC(from eq(1))
It means D is middle point of BC. So, vertices of D will be (3,3,3).
So, vector AD will be 2iˆ+ 2jˆ+ kˆ.

D1 = a+b
D2 = a-b
D1 = 3i + 0j + 0k
D2 = i + 2j + 2k
|D1| = 3
D1.D2 = |D1| . |D2| . cos θ
3 + 0 + 0 = (3) . (3) cos θ 
3 = 9 cosθ
cos^-1 = (⅓)

Δ = 1/2(a * b) = 1/2(b*c) = 1/2(c*a)
2Δ = a*b = b*c = c*a
unit vector = 1/2Δ[a*b + b*c + c*a]

Matrix {(1,2,-1) (1,-1,0) (1,-1,-1)} [a b c]
= {1(1) -2(-1) -1(0)} [a b c]
= 3[a b c]

A * X = B
(A * X)*A = B * A
=> -A(x.A) + x(A.A) = B * A
=> -Ac + x|A|² = B * A
x|A|² = B * A + Ac
x = [B * A + Ac]/|A|²

xa + yb + zc = 0 
We have to prove x = y = z = 0
a,b,c are non planner
x(a-b) +y(b-c) +c(c-a) = 0
Let x = 1, y = 1, z = 1
So, we get a - b + b - c + c - a = 0

|A| = 3, |B| = 4, |C| = 5
Since A.(B + C) = B.(C+A) = C(A+B) = 0...........(1)
|A+B+C|² = |A| + |B|² + |C|² + 2(A.B + B+C + C.A)
= 9+16+25+0
from eq(1) {A.B + B+C + C.A = 0}
therefore, |A+B+C|² = 50
=> |A+B+C| = 5(2)^1/2

Vi=[a→,b→,c→]
(a→ + b→)(b→ + c→)(a→ + c→)
Vf=[(a→ + b→)(b→ + c→)(a→ + c→)]
=(a→ + b→)⋅[(b→ + c→)⋅(a→ + c→)]
=(a→ +b→)[b→⋅a→ + c→⋅a→ + b→⋅c→ + c→⋅c→]
=[b→ c→ a→]+[a→ b→ c→]
Vf=2[a→ b→ c→]
Vf=2Vi.

 (a→*b→)*c→ = (1/3)|b→||c→||a→| 
⇒ − c→*(a→*b→) = (1/3)|b→||c→||a→|
⇒(c→.a→)b→ −(c→.b→)a→ =(1/3)|b→||c→|a→|
Now, as all of them are non-collinear, (c→⋅a→) can be 0 that means,
(c→.b→) = (−1/3)|b→||c→|
⇒|b→|.|c→|cosθ = (−1/3)|b→||c→|
⇒ cosθ = −1/3
sinθ = √1−(1/3)^2
= √8/9 
= (2√2)/√3

The length of median through A = vector(AB + BC)/2 
= (3i + 4k + 5i - 2j + 4k)/2 
= (8i - 2j + 8k)/2 
= 4i - j + 4k 
Length  = √(16 + 1 + 16) = √33

∣u × v∣=∣(a − b) × (a + b)∣
=2∣a × b∣      (∵a × a = b × b = 0)
and ∣a × b∣² + (a ⋅ b)²
=(ab sinθ)² + (abcosθ)²
=a²b² 
⇒∣a×b∣ = a²b²−(a ⋅ b)²
So, ∣u × v∣ = 2∣a × b∣
=2[a ^ 2b² − (a ⋅ b)²]^1/2
=2[(2)²(2)²−(a⋅b)²]^1/2
=2(16−(a⋅b)²)^1/2
 ∴∣a∣=∣b∣=2

According to the given conditions,
(c1)²+c(2)²+(c3)²2 =1,a⋅c=0,b⋅c=0
and cosπ/6 = [(3)^1/2]/2
​(a1b1+a2b2+a3b3]/[(a1)²+(a2)² + (a3)^2)]^1/2 [(b1)^2 + (b2)² + (b3)²]^1/2
Thus a1c1+a2c2+a3c3=0, b1c1+b2c2+b3c3=0
and [(3)^1/2]/2[(a1)²+(a2)² + (a3)²)^1/2 ((b1)² + (b2)² +(b3)²]^1/2
 =a1b1+a2b2 +a3b3
[(a1)²+(a2)² + (a3)²] -(a1b1 + a2b2 + a3b3)²
[(a1)²+(a2)² + (a3)²] [(b1)²+(b2)² + (b3)²]
= -3/4 [(a1)²+(a2)² + (a3)²] [(b1)²+(b2)² + (b3)²]
= 1/4 [(a1)²+(a2)² + (a3)²] [(b1)²+(b2)² + (b3)²]

Let A (0,0); B (4m , 0) and C(4p , 4q)
M1 (2m + 2p, 2q)
M2 (2p , 2q) and M3 (2m , 0)
 Let E , F and G be the point on the median.
E = (2m + 2p) / 4, 2q / 4) = ((m + p) / 2, q / 2)
F = ((2p + 12m) / 4, (2q + 0) / 4) = ((p + 6m) / 2, q / 2)
G = ((2m + 12p) / 4, (0 + 12 q) / 4) = ((m + 6p ) / 2, 3q)
Area of traingle ABC = 1/2
Area of traingle ABC = 1/2 {(0,0,0) (4m,0,1) (4p,4q,1)}
=1/2(16) = 8 unit
Area of triangle EFG = 1/2 {((m + p) / 2, q / 2, 1)) ((p + 6m) / 2, q / 2, 1) ((m + 6p)/2, 3q, 1)}
= 25/8 unit
ar (EFG) /ar (ABC) = {25 / 8} / 8
= 25/64