A vector has magnitude same as that of $\vec{A}=3 \hat{i}+4 \hat{j}$ and is parallel to $\vec{B}=4 \

Question

A vector has magnitude same as that of $\vec{A}=3 \hat{i}+4 \hat{j}$ and is parallel to $\vec{B}=4 \hat{i}+3 \hat{j}$ ?The $x$ and $y$ components of this vector in first quadrant are $x$ and 3 respectively?where $x=$ $\qquad$ ?\

Professor PC Thomas Classes · Answer

(4) : $\vec{A}=3 \hat{i}+4 \hat{j} ; \vec{B}=4 \hat{i}+3 \hat{j}$\ $|\vec{C}|=|\vec{A}|=\sqrt{(3)^{2}+(4)^{2}} \Rightarrow|\vec{C}|=\sqrt{25}$\ $\Rightarrow|\vec{C}|=5$\ $5=\sqrt{x^{2}+3^{2}}$\ $25=x^{2}+3^{2} \Rightarrow x^{2}=25-9=16 \Rightarrow x=4$\ Hence, the vector $C$ is $4 \hat{i}+3 \hat{j}$.