91Ó°ÊÓ

Vector calculation can be used to find the dot product, cross product, addition and subtraction between two vectors as well as the projection of one vector along another vector.The addition and subtraction of vectors give another vector quantity. The cross product of two vectors results in a vector quantity whereas the dot product of two vectors produces a scalar quantity.

The formula for the addition of the vectors is,

$\stackrel{→}{a}+\stackrel{→}{b}=\left(a_x+b_x\right)\hat{i}+\left(a_y+b_y\right)\hat{j}$ (i)

The equation for the dot product of the two vectors is,

$\stackrel{→}{a}·\stackrel{→}{b}=\left(a_x{b}_x+a_y+b_y\right)$ (ii)

The equation for the cross product of the two vectors is,

$\stackrel{→}{a}×\stackrel{→}{b}=\left(a_y{b}_z-b_y+a_z\right)\hat{i}+\left(a_z{b}_x-b_z+a_x\right)\hat{j}+\left(a_x{b}_y-b_x+a_y\right)\hat{k}$ (iii)

The component of along the direction of is,

$\stackrel{→}{b}=\stackrel{→}{a}·\hat{b}$ (iv)

where,$\hat{b}=\frac{\stackrel{→}{b}}{\left|\stackrel{→}{b}\right|}$

It is given that,

role="math" localid="1656307267634" $\stackrel{→}{a}=\left(3.0\right)\hat{i}+\left(5.0\right)\hat{j}$

And,

role="math" localid="1656307300696" $\stackrel{→}{a}=\left(2.0\right)\hat{i}+\left(4.0\right)\hat{j}$

The given vectors are $\stackrel{→}{a}=\left(3.0\right)\hat{i}+\left(5.0\right)\hat{j}$and $\stackrel{→}{a}=\left(2.0\right)\hat{i}+\left(4.0\right)\hat{j}$.

There are no components in the $\hat{k}$direction. Therefore, the equation (iii) can be written as,

$\stackrel{→}{a}×\stackrel{→}{b}=\left(a_x{b}_y-b_x{a}_y\right)\hat{k}$ (v)

Substitute the values of x and y components in equation (v).

$$\begin{gathered} \stackrel{→}{a}×\stackrel{→}{b}=\left[\left(3.0\right)\left(4.0\right)-\left(5.0\right)\left(2.0\right)\right]\hat{k} \\ =2.0\hat{k} \end{gathered}$$

Thus, the cross product, $\stackrel{→}{a}×\stackrel{→}{b}\mathrm{is}2.0\hat{k}$

Substitute the x and ycomponents of $\stackrel{→}{a}$and $\stackrel{→}{b}$in equation (ii),

$$\begin{gathered} \stackrel{→}{a}·\stackrel{→}{b}=\left(3×2+5×4\right) \\ =26 \end{gathered}$$

Thus, the dot product of role="math" localid="1656307675171" $\stackrel{→}{a}$and $\stackrel{→}{b}$is 26 .

Calculate the value of $\left(\stackrel{→}{a}+\stackrel{→}{b}\right)$.

$$\begin{gathered} \left(\stackrel{→}{a}+\stackrel{→}{b}\right)=\left(a_x+b_x\right)\hat{i}+\left(a_y+b_y\right)\hat{j} \\ =\left(5.0\right)\hat{i}+\left(9.0\right)\hat{j} \end{gathered}$$

Now, calculate the dot product of $\left(\stackrel{→}{a}+\stackrel{→}{b}\right)$with $\stackrel{→}{b}$.

So,

$$\begin{gathered} \left(\stackrel{→}{a}+\stackrel{→}{b}\right).\stackrel{→}{b}=\left(5×2\right)+\left(9×4\right) \\ =46 \end{gathered}$$

The dot product of $\left(\stackrel{→}{a}+\stackrel{→}{b}\right)$with $\stackrel{→}{b}$is 46 .

First calculate the unit vector, $\hat{b}$.

role="math" localid="1656308194675" $$\begin{gathered} \hat{b}=\frac{\stackrel{→}{b}}{\left|\stackrel{→}{b}\right|} \\ =\frac{\left(2.0\right)\hat{i}+\left(4.0\right)\hat{j}}{\sqrt{2^2+4^2}} \end{gathered}$$

Now, substitute the above value in equation (iv).

$$\begin{gathered} \stackrel{→}{a}.\hat{b}=\frac{\left(2.0\right)\left(3.0\right)+\left(4.0\right)\left(5.0\right)}{\sqrt{2^2+4^2}} \\ =5.8 \end{gathered}$$

Therefore, the component of $\stackrel{→}{a}$along the direction of $\stackrel{→}{b}$is 5.8 .