A force of magnitude 20 N acts in the direction 4 hatI-3 hatj The force hatF is.

Question

A force of magnitude 20 N acts in the direction 4 hatI-3 hatj The force hatF is. (A) 8i-10 hatj (B) (20/√5)( hati- hatj) (C) (20/√5)( hati+ hatj) (D) 16i-12 hatj

Tardigrade Admin · Accepted Answer

Let barp = 4 hati - 3 hatj then the required vector can be barF = 20 hatp = 20 ((4 hati - 3 hatj)/(5)) = 16 hati - 12 hatj N

Q. A force of magnitude 20 N acts in the direction $4 \hat{I} - 3 \hat{j}$ The force $\hat{F}$ is.

$8 i - 10 \hat{j}$

$\frac{20}{5} (\hat{i} - \hat{j})$

$\frac{20}{5} (\hat{i} + \hat{j})$

$16 i - 12 \hat{j}$

Let $\overset{p}{ˉ} = 4 \hat{i} - 3 \hat{j}$ then the required vector can be
$\overset{ˉ}{F} = 20 \overset{p}{^} = 20 \frac{( 4 i ^ - 3 j ^ )}{( 5 )} = 16 \hat{i} - 12 \hat{j} N$

Q. A force of magnitude 20 N acts in the direction 4I^−3j^​ The force F^ is.

8i−10j^​

5​20​(i^−j^​)

5​20​(i^+j^​)

16i−12j^​