Home
Class 11
PHYSICS
If vec(v)(1)+vec(v)(2) is perpendicular ...

If `vec(v)_(1)+vec(v)_(2)` is perpendicular to `vec(v)_(1)-vec(v)_(2)`, then

A

`vec(v)_(1)` is perpendicular to `vec(v)_(2)`.

B

`|vec(v)_(1)|=|vec(v)_(2)|`

C

`vec(v)_(1)` is a null vector

D

The angle between `vec(v)_(1)` and `vec(v)_(2)` can have any value

Text Solution

Verified by Experts

The correct Answer is:
B, D

If two vectors are normal to each other, then their dot product is zero.
`(vec(v)_(1)+vec(v)_(2)).(vec(v)_(1)-vec(v)_(2))=0 rArrv_(1)^(2)-v_(2)^(2)=0`
`rArr v_(1)^(2)-v_(2)^(2)rArr v_(1)=v_(2)` or `|vec(v)_(1)|=|vec(v)_(2)|`
Promotional Banner

Topper's Solved these Questions

  • VECTORS

    CENGAGE PHYSICS|Exercise Exercise Single Correct|51 Videos
  • TRAVELLING WAVES

    CENGAGE PHYSICS|Exercise Integer|9 Videos
  • WORK, POWER & ENERGY

    CENGAGE PHYSICS|Exercise Archives (integer)|4 Videos

Similar Questions

Explore conceptually related problems

A moving particle of mass m collides elastically with a stationary particle of mass 2m . After collision the two particles move with velocity vec(v)_(1) and vec(v)_(2) respectively. Prove that vec(v)_(2) is perpendicular to (2 vec(v)_(1) + vec(v)_(2))

If a charged particle goes unaccelerated in a region containing electric and magnetic fields, (i) vec(E) must be perpendicular to vec(B) (ii) vec(v) must be perpendicular to vec(E) (iii) vec(v) must be perpendicular to vec(B) (iv) E must be equal to v B

If |vec(V)_(1)+vec(V)_(2)|=|vec(V)_(1)-vec(V)_(2)|and V_(2) is finite, then

Two particles of masses m_(1) and m_(2) in projectile motion have velocities vec(v)_(1) and vec(v)_(2) , respectively , at time t = 0 . They collide at time t_(0) . Their velocities become vec(v')_(1) and vec(v')_(2) at time 2 t_(0) while still moving in air. The value of |(m_(1) vec(v')_(1) + m_(2) vec(v')_(2)) - (m_(1) vec(v)_(1) + m_(2) vec(v)_(2))|

If vec V_(1)=hat i+hat j+hat k;vec V_(2)=ahat i+bhat j+chat k where a,b,c in{-2,-1,0,1,2} then the number of non-zero vectors vec V_(2) which perpendicular to (vec V_(1))/(18) is

A charged particle goes undeflected in a region containing electric and magnetic field. It is possible that (i) vec(E) || vec(B). vec(v) || vec(E) (ii) vec(E) is not parallel to vec(B) (iii) vec(v) || vec(B) but vec(E) is not parallel to vec(B) (iv) vec(E)||vec(B) but vec(v) is not parallel to vec(E)

A charged particle with charge q enters a region of constant, uniform and mututally orthogonal fields vec(E) and vec(B) with a velocity vec(v) perpendicular to both vec(E) and vec(B) , and comes out without any change in magnitude or direction of vec(v) . Then

If vec a+vec b is perpendicular to vec b and vec a+vec 2b is perpendicular to vec a then

Two vector vec(V)andvec(V) have equal magnitudes. If magnitude of vec(A)+vec(B) is equal to n time the magnitude of vec(A)-vec(B) , then angel to between vec(A) and vec(B) is

Find 3 -dimensional vectors vec v_ (1), vec v_ (2), vec v_ (3) satisfying vec v_ (1) * vec v_ (1) = 4, vec v_ (1) * vec v_ (2) = - 2, vec v_ (1) * vec v_ (3) = 6, vec v_ (2) * vec v_ (2) = 2, vec v_ (2) * vec v_ (3) = - 5, vec v_ (3) * vec v_ (3)