Question

If A  3ˆi  4ˆj and B  7ˆi  24ˆj, the vector having the same magnitude as B and parallel to A is

Answer 1

Answer

Given -

$\sf \vec{A} = 3 \hat{\imath} + 4 \hat{\jmath}$

$\sf \vec{B} = 7 \hat{\imath} + 24\hat{\jmath}$

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To find -

Vector having same magnitude as B and parallel to A

$\sf\hat{A} \times |\vec{B}|$

where

$\implies$$\sf\hat{A}$ is the unit vector in direction of a

$\implies$$\sf|\vec{B}|$is magnitude of $\vec{B}$

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Solution -

As the vector is having same magnitude as B -

$|\vec{B}| = \sf\sqrt{{7}^{2} + {24}^{2}}$

$\implies$$\sf \sqrt{625} = 25$

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For same direction along $\vec{A}$ We have to find unit vector in direction of a

Unit vector in direction of A =$\sf \hat A = \frac{\vec A}{|\vec A|}$

$\sf| \vec A| = \sqrt{ {3}^{2} + {4}^{2} } = 5$

$\implies$$\sf\hat A = \frac{ 3 \hat{\imath} + 4 \hat{\jmath}}{5}$

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Required vector =

$\sf \hat{A} \times |\vec{B}|</p><p>= 25 \times \frac{ 3 \hat{\imath} + 4 \hat{\jmath}}{5}$

$\sf= 5(3 \hat{\imath} + 4 \hat{\jmath})$

$\implies$$\sf= 15\hat{\imath} + 20\hat{\jmath}$

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Verification -

$\sf \sqrt{ {15}^{2} + {20}^{2} }$

$\implies$$\sf\sqrt{625} = 25$

Hence the vector having the same magnitude as B and parallel to A is $\sf= 15\hat{\imath} + 20\hat{\jmath}$

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Thanks