What is unit vector along icap + jcap
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4
The dot product gives you the magnitude of the component.
Here (3i+4j).(i+j)= 7
If we want to find out the component vector we have to multiply the magnitude with unit vector i.e i+j/√2
So, required component vector is 7×(i+j)/√2
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Here (3i+4j).(i+j)= 7
If we want to find out the component vector we have to multiply the magnitude with unit vector i.e i+j/√2
So, required component vector is 7×(i+j)/√2
plz mark me as a brainlist
Answered by
7
the answer of the question is- 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 hope its help you plz mark me brainliest alsi
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