Question

find a unit vector in the direction of the vector 3i+4j
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Answer 1

Explanation:

let V= 3i+4j

magnitude of v=|v|=

$\sqrt{3 {}^{2} } + \sqrt{4 {}^{2} }$

$\sqrt{25}$

=5

v=x|v| where is the unit vector along v

x=v/|v|=3/5i+4/5j

pg no 21

physics textbook

11th standard

Answer 2

Given: A vector 3i+4j

To find: A unit vector in direction of 3i+4j

Solution: A unit vector is a vector whose magnitude is equal to 1.

For finding a unit vector in direction of 3i+4j, we divide the vector 3i+4j by its magnitude.

Magnitude of vector 3i+4j is equal to

$= \sqrt{ {3}^{2} + {4}^{2} }$

$= \sqrt{9 + 16}$

$= \sqrt{25}$

$= 5$

Therefore, unit vector

$= \frac{3i + 4j}{5}$

$= \frac{3}{5} i + \frac{4}{5} j$

Therefore, unit vector in direction of 3i+4j is $\frac{3}{5} i + \frac{4}{5} j.$