Question

The resultant of two vectors A and B is vector C=2.2i+3.4j. If vector A=1.5i-2.0j, find the magnitude of B and the angle it makes with the positive x-​

Answer 1

$\underline{ \bold{question: }}$

$\sf{The \: resultant \: of \: two \: vectors \: A \: and \: B \: is \: vector }\\ \sf{ C=2.2 \hat{i}+3.4 \hat{j}. \: If \: vector \: A=1.5 \hat{i}-2.0 \hat{j}, find} \\ \sf{ the \: magnitude \: of \: B \: and \: the \: angle \: it \: makes \: with} \\ \sf{ \: the \: positive \: x-axis. \qquad \qquad\qquad \qquad \qquad \ \ \ \ \ }$

$\underline{ \bold{solution : }}$

we have,

$\sf \vec{A}=1.5 \hat{i}-2.0 \hat{j}$

and,

$\sf \vec{C}=2.2 \hat{i}+3.4 \hat{j}$

Thus ,

$\begin{aligned} \sf \vec{B}& = \sf \vec{C} - \vec{A} \\ \\ & = (2.2 \hat i + 3.4 \hat j) - (1.5 \hat i - 2.0 \hat{j}) \\ \\ & = (2.2 - 1.5) \hat{i} + (3.4 + 2.0) \hat{j} \\ \\ & = 0.7 \hat{i} + 5.4 \hat{j} \end{aligned}$

Thus ,

magnitude of vector B ,

$\begin{aligned} \sf | \vec{B}| & = \sf \sqrt{ {(0.7)}^{2} + {(5.4)}^{2} } \\ \\ & = \sqrt{0.49 + 29.16 } \\ \\ & = \sqrt{29.65} \\ \\ & = \ \ 5.445 \end{aligned}$

Now , let theta $\theta$ be the angle made by vector B with the x-axis.

Thus ,

$\begin{aligned} \tan \theta & = \sf \frac{ \small{vertical \: compoment}}{ \small{horizontal \: component}} \\ \\ & = \frac{5.4}{0.7} \\ \\ & = \frac{54}{7} \\ \\ \implies \quad \theta \ & = { \tan}^{ - 1} \bigg( \frac{54}{7} \bigg) \end{aligned}$