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

The correct option is D 20 From the diagram, ABC is right angle triangle where lt;A=90^o, #160;AC=10 and AB=r #160;So, cos60^0=ACBC ...

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Standard XII

Physics

Multiplication with Vectors

A displacemen...

Question

A displacement vector, at an angle of $30^{\circ}$ with y-axis has a $x$ -component of $10$ units. Then the magnitude of the vector is

5.0

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Solution

The correct option is D $20$
From the diagram, ABC is right angle triangle where $< A = 90^{o}, A C = 10$ and $A B = r$
So, $cos 60^{0} = \frac{A C}{B C}$

$\Rightarrow \frac{1}{2} = \frac{10}{r}$

$\Rightarrow r = 20$ units

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

The correct option is D 20From the diagram, ABC is right angle triangle where <A=90o,AC=10 and AB=r So, cos600=ACBC⇒12=10r ⇒r=20 units

A displacement vector, at an angle of $30^{\circ}$ with y-axis has a $x$ -component of $10$ units. Then the magnitude of the vector is

The correct option is D $20$
From the diagram, ABC is right angle triangle where $< A = 90^{o}, A C = 10$ and $A B = r$
So, $cos 60^{0} = \frac{A C}{B C}$

$\Rightarrow \frac{1}{2} = \frac{10}{r}$

$\Rightarrow r = 20$ units