Question

I and j are unit vectors along x and y axis respectively.what is the magnitude and direction of the vectors i+j and i-j?whata re thr

Answer 1

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Answer 2

Answer: The magnitude of both vectors is $|\vec{A}| = \sqrt{2}$ and $|\vec{B}| = \sqrt{2}$ and direction is $\theta = \dfrac{\pi}{4}$ and $\theta = \dfrac{-\pi}{4}$ respectively

Explanation:

Let us consider two vectors A and B.

$\vec{A} = \hat i+\hat j$

$\vec{B} = \hat i-\hat j$

The magnitude of $\vec{A}$

$|\vec{A}|= \sqrt{A_{x}^{2}+A_{y}^{2}}$

$|\vec{A}|= \sqrt{1^{2}+1^{2}}$

$|\vec{A}| = \sqrt{2}$

Now, the direction of the vector A

$\theta = tan^{-1}(\dfrac{y}{x})$

Where, $\theta$ ie the direction of vector

$\theta = tan^{-1}(1)$

$\theta = \dfrac{\pi}{4}$

Now, the magnitude of  $\vec{B}$

$|\vec{B}|= \sqrt{B_{x}^{2}+B_{y}^{2}}$

$|\vec{B}|= \sqrt{1^{2}+1^{2}}$

$|\vec{B}| = \sqrt{2}$

Now, the direction of the vector B

$\theta = tan^{-1}(\dfrac{y}{x})$

Where, $\theta$ ie the direction of vector

$\theta = tan^{-1}(-1)$

$\theta = \dfrac{-\pi}{4}$

Hence, the magnitude of both vectors is $|\vec{A}| = \sqrt{2}$ and $|\vec{B}| = \sqrt{2}$ and direction is $\theta = \dfrac{\pi}{4}$ and $\theta = \dfrac{-\pi}{4}$ respectively