Question #2cd38

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Question #2cd38

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Answer 1 · 2018-01-17T13:38:10

As I understood

Explanation:

Let $\to A = A_{x} ˆ i + A_{y} ˆ j + A_{z} ˆ k$ $vecA=A_xhati+A_yhatj+A_zhatk$ and
$\to B = B_{x} ˆ i + B_{y} ˆ j + B_{z} ˆ k$ $vecB=B_xhati+B_yhatj+B_zhatk$

To find ( $\to A - \to B$ $vecA-vecB$ )
Inserting values of both the vectors in terms of unit vectors we get

$\to A - \to B = (A_{x} ˆ i + A_{y} ˆ j + A_{z} ˆ k) - (B_{x} ˆ i + B_{y} ˆ j + B_{z} ˆ k)$ $vecA-vecB=(A_xhati+A_yhatj+A_zhatk)-(B_xhati+B_yhatj+B_zhatk)$
$\Rightarrow \to A - \to B = (A_{x} - B_{x}) ˆ i + (A_{y} - B_{y}) ˆ j + (A_{z} - B_{z}) ˆ k$ $=>vecA-vecB=(A_x-B_x)hati+(A_y-B_y)hatj+(A_z-B_z)hatk$

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Question #2cd38

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