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If the vector is collinear with the vector (2√2, 14) and = 10, then
a = 2(2)^{½ }i  j + 4k
b = 10
[(2(2λ)^{2} + (1λ)^{2} + (4λ)^{2}]^{½}
=[ 8λ^{2} + λ^{2} + 16λ^{2}]^{½} = 10
= 25λ^{2} = (10)^{2}
25λ^{2} = 100
λ = +2
b = +(4(2)^{½} i  2j + 8k)
Therefore, 2a + b
The vertices of a triangle are A(1, 1, 2), B(4, 3, 1) and C(2, 3, 5). A vector representing the internal bisector of the angle A is
Let AD is the bisector of ∠A.Then,
BD/DC = AB/ACeq(1)
Given the vertices of triangle,
AB = √3^{2} + 2^{2} + 1^{2} = √14
AC = √1^{2} + 2^{2} + 3^{2} = √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.
Angle between diagonals of a parallelogram whose side are represented by
D1 = a+b
D2 = ab
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)
Vector make an angle θ = 2π/3. if , then is equal to
Unit vector perpendicular to the plane of the triangle ABC with position vectors of the vertices
A, B, C is
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]
If are two noncollinear vectors such that , then is equal to
Vector of length 3 unit which is perpendicular to and lies in the plane of and
A * X = B
(A * X)*A = B * A
=> A(x.A) + x(A.A) = B * A
=> Ac + xA^{2} = B * A
xA^{2} = B * A + Ac
x = [B * A + Ac]/A^{2}
If ^{a ,b,c} are linearly independent vectors, then which one of the following set of vectors is linearly dependent ?
xa + yb + zc = 0
We have to prove x = y = z = 0
a,b,c are non planner
x(ab) +y(bc) +c(ca) = 0
Let x = 1, y = 1, z = 1
So, we get a  b + b  c + c  a = 0
Let be vectors of length 3,4,5 respectively. Let be perpendicular to , and . then
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}
Given the vertices A (2, 3, 1), B(4, 1, –2), C(6, 3, 7) & D(–5, –4, 8) of a tetrahedron. The length of the altitude drawn from the vertex D is
for a non zero vector If the equations hold simultaneously, then
The volume of the parallelopiped constructed on the diagonals of the faces of the given rectangular parallelopiped is m times the volume of the given parallelopiped. Then m is equal to
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.
If u and v are unit vectors and θ is the acute angle between them, then 2u × 3v is a unit vector for
The value of a, for which the points A,B,C with position vectors and respectively are the vertices of a right angled triangle with C = π/2 are
A particle is acted upon by constant forces which displace ot from a point to the point . The workdone in standard units by the force is given by
If are noncoplaner vectors and λ is a real number, then the vectors are noncoplaner for
Let be non zero vectors such that , If θ is the acute angle between the vectors , then sin θ equals is
(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 noncollinear, (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 vectors are the sides of a triangle ABC. The length of the median through A is
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}
=a^{2}b^{2}
⇒∣a×b∣ = a^{2}b^{2}−(a ⋅ b)^{2}
So, ∣u × v∣ = 2∣a × b∣
=2[a ^ 2b^{2 }− (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
Let and be three nonzero vectors such that is a unit vector perpendicular to both . if the angle between is π/6, then is equal to
According to the given conditions,
(c1)^{2}+c(2)^{2}+(c3)^{2}2 =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}]
A point taken on each median of a triangle divides the median in the ratio 1 : 3, reckoning from the vertex.
Then the ratio of the area of the triangle with vertices at these points to that of the original triangle is
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
209 videos218 docs139 tests

209 videos218 docs139 tests
