Answer: Option C. -> I tried and did not get the answer initially but after a few attempts I think I have the right answer now.
:
C

Answer: Option C. -> −3√3
:
C
Well normally I guess by now you know how to find component of a vector along a given vector i.e.
(Magnitude of vector)×(cosine of angle between the two vectors)
This takes care of everything
$\Rightarrow$ In this case
$6\times cos{150}^{\circ }=6\times cos(90+60)$
= -$6sin{60}^{\circ }=-6\times \frac{\sqrt{3}}{2}=-3\sqrt{3}$
Caution:- Component of a vector is scalar. I agree it's tough to understand and remember trigonometric
identitiesintuitively. Also component of a vector is just the measure of the shadow
$\Rightarrow$

Here the length of shadow is $6cos{30}^{\circ }=6\times \frac{3}{2}=3\sqrt{3}$ but this component is on -ve x-axis.
But since the question asks the component along the direction of +vex-axis, so the component should be $-3\sqrt{3}$

Answer: Option C. -> π2
:
C
Let $\theta$ be the angle between A and B. Then magnitudes of vectors P and Q are given by
$P^2=A^2+B^2+2AB\cos \theta$
and $Q^2=A^2+B^2–2AB\cos \theta$
Since P = Q, it follows that $\cos \theta =0or\theta =\frac{\pi }{2}.$ Hence correct choice is (c).

Answer: Option D. -> zero 
:
D
The magnitude of $\vec{A}\times \vec{B}=AB\sin \theta =3\times 4\times \sin {180}^{\circ }=0.$
Hence the correct choice is (d)

Answer: Option B. -> B is a null vector
:
B
Given $\vec{A}+\vec{B}=\vec{A}–\vec{B}$
Which gives $\vec{B}=–\vec{B}$. This is possible only if $\vec{B}$ is a null vector.
Hence the correct choice is (b).

Answer: Option D. -> 180∘
:
D
Since $\vec{A}\times \vec{B}=–\vec{B}\times \vec{A},\vec{C}=–\vec{D},$ i.e. vectors $\vec{C}and\vec{D}$ are in opposite directions.
Hence the correct choice is (d).

Answer: Option B. -> 5 in North East direction
:
B
Watch the next video for solution.

Answer: Option D. -> I must swim somewhere between right and straight
:
D
You must swim somewhere between right and straight.

Answer: Option B. -> False
:
B
If a vector is rotated by any angle which is not a multiple of ${360}^{\circ }$or $2\pi$radians, then it is different. But if rotated through a multiple of ${360}^{\circ }$, it remains the same. As it is simply making full circles, the direction will be the same as the initial direction.

Answer: Option B. -> (A) & (D)
:
B
From the diagram, both A and D have a magnitude of 2 units and are facing opposite directions.