Question

The component of vector A^=2i^+3j^ along the X-axis is​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

Let , \vec{a}=\hat{i}+\hat{j}

Magnitude of \vec{a}=\sqrt{1^2+1^2}=\sqrt{2}

Let \theta is the angle made by \vec{a} with x-axis

Then, value of \theta=tan^{-1}\frac{1}{1}=tan^{-1}1=45^{\circ} , direction is shown in figure of \vec{a}

Similarly, Let \vec{b}=\hat{i}-\hat{j}

Magnitude of \vec{b}=\sqrt{1^2+(-1/^2}=\sqrt{2}

angle made by \vec{b} with x-axis is tan^{-1}\frac{-1}{1}=tan^{-1}(-1)=-45^{\circ}, direction is shown in figure .

Now, Let \vec{A}=2\hat{i}+3\hat{j}

The components of a vector A= 2i + 3j along the directions of (\hat{i}+\hat{j}) : \vec{A}(2,3)

The components of a vector A = 2i + 3j along the directions of (\hat{i}-\hat{j}) : \vec{A}(2,-3)