A displacement vector- at an angle of 30- with y-axis has an x-component of 10 units- Then the magnitude of the vector is-

Vector A has magnitude of 5 units and makes an angle of $30^{\circ}$ with the x-axis. Vector B has magnitude 10 units & make an angle of $60^{\circ}$ with the x-axis. Find $\to A + \to B$

Q. A vector has a magnitude

12

. When its tail is at origin, it lies between the positive x - axis and the negative y - axis and makes an angle

30^{\circ}

with the x- axis. It’s y - component is

x

component of

¯ ¯ ¯ a

is twice of its

y

-component. If the magnitude of the vector is

5 \sqrt{2}

and it makes an angle of

135^{\circ}

with

z

-axis then the vector is

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A displacement vector, at an angle of 30∘ with y-axis has an x-component of 10units. Then the magnitude of the vector is:

The correct option is C 20Angle with y-axis is 300, thus angle with x-axis is 600. Hence its x-component is Vcos600=V/2, which is given as 10.Thus V2=10 or V=20units.

A displacement vector, at an angle of $30^{\circ}$ with y-axis has an x-component of $10 u n i t s$ . Then the magnitude of the vector is:

The correct option is C $20$
Angle with y-axis is $30^{0}$ , thus angle with x-axis is $60^{0}$ . Hence its x-component is $V c o s 60^{0} = V / 2$ , which is given as $10$ .
Thus $\frac{V}{2} = 10$ or $V = 20 u n i t s .$