Question

if $\vec A = 2\vec i + \vec j$ and $\vec B = \vec i - \vec j$, sketch vectors graphically and find the component of $\vec A$ along $\vec B$ and perpendicular to $\vec B$.
(A) Component of A along B: $\dfrac{1}{2}\left( {\vec i - \vec j} \right)$
      Component of A perpendicular to B: $\dfrac{4}{2}\left( {\vec i + \vec j} \right)$
(B) Component of A along B: $\dfrac{1}{2}\left( {\vec i - \vec j} \right)$
      Component of A perpendicular to B: $\dfrac{3}{2}\left( {\vec i + \vec j} \right)$
(C) Component of A along B: $\dfrac{1}{2}\left( {\vec i - \vec j} \right)$
      Component of A perpendicular to B: $\dfrac{1}{2}\left( {\vec i + \vec j} \right)$
(D) Component of A along B: $\dfrac{1}{3}\left( {\vec i - \vec j} \right)$
       Component of A perpendicular to B: $\dfrac{3}{2}\left( {\vec i + \vec j} \right)$

Answer 1

Hint: For the diagram's construction, use the component of $\vec j$ on the y-axis and use the component of $\vec i$ on the x-axis. To calculate the components of $\vec A$, first, perform the dot product of the two given vectors and then use the dot product's value in the formula, so that calculation becomes easy.

Complete step by step answer:
By using the information of the given vectors, we will draw the vectors graphically as,
Now we will do the dot product of two given vectors.

Therefore we get

\[\begin{array}{l}
\vec A.\vec B = \left( {2\vec i + \vec j} \right).\left( {\vec i - \vec j} \right)\\
\vec A.\vec B = \left( 2 \right)\left( 1 \right) + \left( 1 \right)\left( { - 1} \right)\\
\vec A.\vec B = 2 - 1\\
\vec A.\vec B = 1
\end{array}\]

Write the formula of the component of $\vec A$ along $\vec B$, therefore we get

${A_b} = \left( {\dfrac{{\vec A.\vec B}}{{{{\left| {\vec B} \right|}^2}}}} \right)\vec B$……. (1)

Now we will calculate the magnitude of the B vector, so we get

$\begin{array}{l}
\left| {\vec B} \right| = \sqrt {{1^2} + {1^2}} \\
\left| {\vec B} \right| = \sqrt 2
\end{array}$

By substituting values of dot product and magnitude of B vector in the equation (1), we can calculate the component of $\vec A$ along $\vec B$,

\[\begin{array}{l}
{A_b} = \left( {\dfrac{{\vec A.\vec B}}{{{{\left| {\vec B} \right|}^2}}}} \right)\vec B\\
{A_b} = \left( {\dfrac{1}{{{{\left( {\sqrt 2 } \right)}^2}}}} \right)\left( {\vec i - \vec j} \right)\\
{A_b} = \dfrac{1}{2}\left( {\vec i - \vec j} \right)
\end{array}\]

To calculate the component of $\vec A$ perpendicular to$\vec B$, we will subtract the value of ${A_b}$ from the A vector.

Therefore, we get

\[\begin{array}{l}
{A_{ \bot B}} = \vec A - {A_B}\\
{A_{ \bot B}} = \left( {2\vec i + \vec j} \right) - \dfrac{1}{2}\left( {\vec i - \vec j} \right)\\
{A_{ \bot B}} = \left( {2 - \dfrac{1}{2}} \right)\vec i + \left( {1 + \dfrac{1}{2}} \right)\vec j\\
{A_{ \bot B}} = \dfrac{3}{2}\left( {\vec i + \vec j} \right)
\end{array}\]

Therefore, if $\vec A = 2\vec i + \vec j$ and$\vec B = \vec i - \vec j$, the components of $\vec A$ along $\vec B$ and perpendicular to $\vec B$ are \[\dfrac{1}{2}\left( {\vec i - \vec j} \right)\] and\[\dfrac{3}{2}\left( {\vec i + \vec j} \right)\]

So, the correct answer is “Option B”.

Note:
Remember the formulas related to the vectors as well revise the concept of vector calculation, like how to do dot product, cross products, addition and subtraction of vectors, because these things are commonly used for this type of questions.