Answer: Option D. -> zero magnitude and no specific direction
:
D
If vectors A and B are equal, i.e. A = B, then their difference (A – B) is defined as a null vector. It has a zeromagnitude and no specific direction.
Hence the correct choice is (d).

Answer: Option B. -> (C) & (B)
:
B
For 2 vectors to be equal their magnitude and direction has to be same.
Option A: says vector A and E surely they are pointed in the same direction but if we see magnitude wise the vector A has a magnitude of 2 units or it is only 2 boxes long, while vector E has magnitude of 3 units, so vector A & E can't be same.
Option B: says vector C & B, they are surely pointing in the same direction are their magnitudes also same.
Vector C has a unit of $2\sqrt{2}$units and so does vector B.
So option B is correct.
Here we conclude that we can translate a vector anywhere on the 3D space, it wouldn't change it as long as I keep the magnitude of the vector same and keep it pointing in the same direction.

Answer: Option A. -> A||B = 3√3;A⊥B=3
:
A

So, we need to resolve the vector A along the direction of B and perpendicular to it.

Angle between A & B is ${30}^{\circ }$
So component along the direction B is using trigonometric ration
$6cos{30}^{\circ }$ = compound of A along B = $3\sqrt{3}$ component of A perpendicular to B = 6 $sin{30}^{\circ }$ = 3

Answer: Option C. -> The magnitude of the vector is doubled and its direction is reversed
:
C
Multiplication of a vector by a real negative number – n makes its magnitude n times and also reverses thedirection of the vector.
Hence the correct choice is (c).

Answer: Option C. -> 120∘
:
C
$R^2=A^2+B^2+2AB\cos \theta$. It is given that $R=A=B.$ Putting these values we have $A^2=A^2+A^2+2A^2\cos \theta$
or $\cos \theta =-\frac{1}{2}$ which gives $\theta ={120}^{\circ }$
Hence the correct choice is (c)

Answer: Option B. -> √2A 
:
B
As shown in figure, the angle $\theta$ between vectors A and B is ${90}^{\circ }$.
Also A = B. Therefore, the magnitude of the resultant is given by

$R^2=A^2+B^2+2AB\cos \theta$
$=A^2+A^2+2A^2\cos {90}^{\circ }$
$=2A^2$
Or $R=\sqrt{2}A.$
Hence the correct choice is (b)

Answer: Option C. -> →A×→B=→C 
:
C
As explained above, choices (a) and (b) are not possible. Since vector C is perpendicular to both vectors A and B,choice (c) is possible. Choice (d) is not possible because vector B is not perpendicular to vector A.

Answer: Option A. -> they have the same magnitude and the same direction.
:
A
Two vectors are said to be equal if they have the same magnitude and direction.
In the figure below, three equal vectors have been represented.

Answer: Option C. -> A||B = -5; A⊥B=5√3
:
C

So component of A along direction of B is $Acos{120}^{\circ }=-Asin30=-5$
$[\because cos(90+\theta )=-sin\theta ]$
Here we can see if just measure the shadow you will get a $cos60=\frac{a}{2}=5$

But since this 5 is in the opposite direction of vector B so we say that the component of A along the
direction B is -5.
Now component in the perpendicular direction of B is
$Asin\left({120}^{\circ }\right)=Acos30=5\sqrt{3}$
$[\because sin(90+\theta )=cos\theta ]$

Just the shadow measurement is a cos ${30}^{\circ }$ and it's in the perpendicular direction, so no need to
change sign.
So component along the direction is always $acos\theta$ and in perpendicular direction is $asin\theta$. This will take care of sign so don't worry.

Answer: Option B. -> A + B = - (B + A)
:
B
Vector addition is commutative, i.e. A + B = B + A. Vector addition is also associative,
i.e. (A + B) + C = A + (B + C) = (A + C) + B.
Hence choice (b) is false.