The unit vector along icap +jcap
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Answered by
378
unit vector along ( i + j) is ( i + j)/| i + j|
|i + j | is magnitude of ( i + j)
we know,
i and j are perpendicular to each other
So, |i + j| = √(i² + j² + 2.I.j.cos90°)
= √( 1 + 1) = √2
hence, unit vector along (i + j) is (i + j)/√2
|i + j | is magnitude of ( i + j)
we know,
i and j are perpendicular to each other
So, |i + j| = √(i² + j² + 2.I.j.cos90°)
= √( 1 + 1) = √2
hence, unit vector along (i + j) is (i + j)/√2
Answered by
77
Answer:
unit vector along ( i + j) is ( i + j)/| i + j|
|i + j | is magnitude of ( i + j)
we know,
i and j are perpendicular to each other
So, |i + j| = √(i² + j² + 2.I.j.cos90°)
= √( 1 + 1) = √2
hence, unit vector along (i + j) is (i + j)/√2
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Explanation:
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