Question #848d3

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Question #848d3

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Answer 1 · 2016-05-21T08:09:26

Given $\to A = A_{x} ˆ i + A_{y} ˆ j + A_{z} ˆ k$ $vec A = A_x hat i + A_y hat j + A_z hat k$

So
$\to A . \to A = (A_{x} ˆ i + A_{y} ˆ j + A_{z} ˆ k) \cdot (A_{x} ˆ i + A_{y} ˆ j + A_{z} ˆ k)$ $vec A.vecA =( A_x hat i + A_y hat j + A_z hat k)*( A_x hat i + A_y hat j + A_z hat k)$

$=>|vec A||.vecA|cos0 =( A_x hat i*A_x hat i + A_y hat j*A_x hat i + A_z hat k*A_x hat i+A_x hat i* A_y hat j + A_y hat j* A_y hat j + A_z hat k* A_y hat j +A_x hat i*A_z hat k + A_y hat j*A_z hat k + A_z hat k*A_z hat k) .....(1)$

We know
$hati*hati=|hati||hati|cos0=1xx1xx1=1$
similarly
$hatj*hatj=hatk*hatk=1$

We know
$hati*hatj=|hati||hatj|cos90=1xx1xx0=0$
similarly
$hati*hatj=hatj*hatk=hatk*hati=0$

So we have from relation (1)

$|vecA|^2=A_x^2 + A_y^2 + A_z^2$

$=>|A| = sqrt(A_x^2 + A_y^2 + A_z^2)$

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Question #848d3

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