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Need solution for RD sharma maths class 12 chapter 22 Algebra of Vector exercise 22.5 question 10

Answers (1)

 

Answer: \frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}

Hint: a unit vector parallel to

 \vec{a}=\hat{a}=\frac{\vec{a}}{\left | \vec{a} \right |}\\ \left | \vec{a} \right |=\sqrt{x^{2}+y^{2}}

Given: \vec{a}=\hat{i}+\sqrt{3}\hat{j}

Solution:

We can find the magnitude of \vec{a}

\left | \vec{a} \right |=\sqrt{x^{2}+y^{2}}

Here x = 1 and  y=\sqrt{3}

\left |\vec{a} \right |=1^{2}+(\sqrt{3})^{2}\\ =\sqrt{1+3}=4\\ \left |\vec{a} \right |=2

Unit vector parallel to \vec{a}=\hat{a}=\frac{\vec{a}}{\left | \vec{a} \right |}\\

=\frac{1}{2}\left ( \hat{i}+\sqrt{3}\hat{j} \right )\\ =\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}

 

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