Vector A- is 2 cm long and is 60o above the x-axis in the first quadrant- Vector B- is 2 cm long and is 60o below the x-axis in the fourth quadrant- The sum A- - B- is a vector of magnitude

Vector $\to A$ is 2 cm long and is $60^{\circ}$ above the +ve X axis. Vector $\to A$ is 2 cm long and is $60^{\circ}$ below the x - axis in the fourth quadrant. The value of $\to A$ + $\to B$ is

Q. Vector

\to A

is of length

2 cm

and makes an angle

60^{\circ}

with the

x

-axis in the first quadrant. Vector

\to B

is of length

2 cm

and

60^{\circ}

below the

x

-axis in the fourth quadrant. The sum

\to A + \to B

is a vector of magnitude in

cm

Q. Two vectors

\to A

and

\to B

of magnitude 2 units and 1 unit, respectively are directed along the x-axis and y-axis. Their resultant

\to A + \to B

is directed along the line:

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Vector →A is 2cm long and is 60o above the x-axis in the first quadrant. Vector →B is 2cm long and is 60o below the x-axis in the fourth quadrant. The sum →A+→B is a vector of magnitude

The correct option is B 2cm along positive x-axisFrom given definition →A=2cos60^i+2sin60^j→B=2cos(−60)^i+2sin(−60)^j=2cos60^i−2sin60^j→A+→B=2×2cos60^i+0^j=4.12^i=2^i∣∣→A+→B∣∣=2.

Vector $\to A$ is $2 c m$ long and is $60^{o}$ above the x-axis in the first quadrant. Vector $\to B$ is $2 c m$ long and is $60^{o}$ below the x-axis in the fourth quadrant. The sum $\to A + \to B$ is a vector of magnitude

The correct option is B $2 c m$ along positive x-axis
From given definition
$\to A = 2 c o s 60^i + 2 s i n 60^j$
$\to B = 2 cos (- 60)^i + 2 sin (- 60)^j = 2 cos 60^i - 2 sin 60^j$
$\to A + \to B = 2 \times 2 cos 60^i + 0^j = 4. \frac{1}{2}^i = 2^i$
$∣ ∣ \to A + \to B ∣ ∣ = 2$ .