If A  3ˆi  4ˆj and B  7ˆi  24ˆj, the vector having the

1.	If A  3ˆi  4ˆj and B  7ˆi  24ˆj, the vector having the same magnitude as B and parallel to A is
Answer» Answer Given - $\sf \vec{A} = 3 \hat{\imath} + 4 \hat{\jmath}$ $\sf \vec{B} = 7 \hat{\imath} + 24\hat{\jmath}$ ━━━━━━━━━━━━━ 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 $\vec{B}$ ━━━━━━━━━━━━━ Solution - As the vector is having same magnitude as B - $\|\vec{B}\| = \sf\sqrt{{7}^{2} + {24}^{2}}$ $\implies$ $\sf \sqrt{625} = 25$ ━━━━━━━━━━━━━ For same direction along $\vec{A}$ We have to find unit vector in direction of a Unit vector in direction of A = $\sf \hat A = \frac{\vec A}{\|\vec A\|}$ $\sf\| \vec A\| = \sqrt{ {3}^{2} + {4}^{2} } = 5$ $\implies$ $\sf\hat A = \frac{ 3 \hat{\imath} + 4 \hat{\jmath}}{5}$ ━━━━━━━━━━━━━ Required vector = $\sf \hat{A} \times \|\vec{B}\|= 25 \times \frac{ 3 \hat{\imath} + 4 \hat{\jmath}}{5}$ $\sf= 5(3 \hat{\imath} + 4 \hat{\jmath})$ $\implies$ $\sf= 15\hat{\imath} + 20\hat{\jmath}$ ━━━━━━━━━━━━━ 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 $\sf= 15\hat{\imath} + 20\hat{\jmath}$ ━━━━━━━━━━━━━ Thanks