Question

Forces of magnitudes 5 and 3 units acting in the directions 6i + 2j + 3k and 3i - 2j + 6krespectively act on a particle which is displaced from the point (2,2, -1) to (4,3,1). Find thework done​

Answer 1

GIVEN :-

• Magnitude of forces = 5 and 3 units.
• direction of forces = 6i + 2j + 3k and 3i - 2j + 6k.

TO FIND :-

• The total work done.

SOLUTION :-

★ direction of force of magnitude 5 units = 6i + 2j + 3k.

★ Direction of force of magnitude 3 units = 3i - 2j + 6k.

★ To Find the Total work done, we have to find the resultant force of vector and displacement vector. Let's start with the resultant force,

★ Let us considered that the resultant force be vector F.

Now,

$\\ \implies \displaystyle \: \tt \: \vec{F} = \bigg( 5 \dfrac{6 \hat{i} + 2 \hat{j} + 3 \hat{k}}{ \sqrt{(6) ^{2} + (2) ^{2} + (3) ^{2} } } \bigg) + \bigg(3 \dfrac{3 \hat{i} - 2 \hat{j} + 6\hat{k}}{ \sqrt{(6) ^{2} + ( - 2) ^{2} + (3) ^{2} } } \bigg)$

$\implies \displaystyle \: \tt \: \vec{F} = \bigg( 5 \dfrac{6 \hat{i} + 2 \hat{j} + 3 \hat{k}}{ \sqrt{36 + 4 + 9} } \bigg) + \bigg(3 \dfrac{3 \hat{i} - 2 \hat{j} + 6\hat{k}}{ \sqrt{36 + 4 + 9} } \bigg)\\ \\ \\ \implies \displaystyle \: \tt \: \vec{F} = \bigg( 5 \dfrac{6 \hat{i} + 2 \hat{j} + 3 \hat{k}}{ \sqrt{ 49} } \bigg) + \bigg(3 \dfrac{3 \hat{i} - 2 \hat{j} + 6\hat{k}}{ \sqrt{49} } \bigg)$

$\implies \displaystyle \: \tt \: \vec{F} = \bigg( 5 \dfrac{6 \hat{i} + 2 \hat{j} + 3 \hat{k}}{ 7 } \bigg) + \bigg(3 \dfrac{3 \hat{i} - 2 \hat{j} + 6\hat{k}}{ 7 } \bigg)$

$\displaystyle \: \tt \: \implies \: \vec{F} = \bigg(\dfrac{30 \hat{i} + 10 \hat{j} + 15 \hat{k}}{7} \bigg) + \bigg(\dfrac{9 \hat{i} - 6 \hat{j} + 18 \hat{k}}{7} \bigg)$

$\displaystyle \: \tt \: \implies \: \vec{F} = \bigg(\dfrac{30 \hat{i} + 10 \hat{j} + 15 \hat{k} + 9 \hat{i} - 6 \hat{j} + 18 \hat{k}}{7} \bigg)$

$\displaystyle \: \tt \: \implies \: \vec{F} = \dfrac{39 \hat{i} + 4 \hat{j} + 33 \hat{k}}{7}$

$\displaystyle \: \bf \: \implies \: \boxed{ \vec{F} = \dfrac{1}{7} \bigg\lgroup39 \hat{i} + 4 \hat{j} + 33 \hat{k} \bigg \rgroup}$

★ Hence the resultant force is $\displaystyle \: \bf \: \: \vec{F} = \dfrac{1}{7} \bigg\lgroup39 \hat{i} + 4 \hat{j} + 33 \hat{k} \bigg \rgroup$

★ Now let the displacement vector be 'b'. Now as we know that forces are displaced from point (2,2,-1) to (4,3,1) so the displacement vector will be,

$\\ \displaystyle \tt \implies \vec{d} = \bigg(4 \hat{i} + 3 \hat{j} + \hat{k} \bigg)- \bigg(2 \hat{i} + 2 \hat{j} - \hat{k} \bigg)$

$\displaystyle \tt \implies \vec{d} = 4 \hat{i} + 3 \hat{j} + \hat{k} - 2 \hat{i} - 2 \hat{j} + \hat{k}$

$\displaystyle \tt \implies \vec{d} = 4 \hat{i} - 2 \hat{i} +3 \hat{j} - 2 \hat{j} + \hat{k} + \hat{k}$

$\displaystyle \bf \: \implies \boxed{ \vec{d} = {2 \hat{i} + \hat{j} + 2 \hat{k}}}$

★ Hence the displacement vector is $\displaystyle \bf \: \bigg \lgroup \vec{d} = {2 \hat{i} + \hat{j} + 2 \hat{k}} \bigg \rgroup$

★ Now we will use the formula of total work done i.e Total work done = Displacement vector × Resultant force.

$\\ \displaystyle \tt \implies \: \vec{d} \times \vec{F}$

$\displaystyle \tt \implies \: \dfrac{1}{7} \bigg\lgroup \big(39 \hat{i} + 4 \hat{j} + 33 \hat{k} \big) \times \big(2 \hat{i} + \hat{j} + 2 \hat{k} \big) \bigg \rgroup$

$\displaystyle \tt \implies \: \dfrac{1}{7} \big(78 + 4 + 66 \big)$

$\displaystyle \tt \implies \: \dfrac{1}{7} (148)$

$\displaystyle \tt \implies \: \boxed{ \red{ \bf \: Total \: work \: done \: = \dfrac{\: 148}{7} units }}$

★ Hence the total work done is 148/7 units.

Answer 2

Answer:

Given :

▪ A vector quantity has been provided.

$\bigstar\:\underline{\boxed{\bf{\red{\vec{A}=-3\hat{i}+2\hat{j}+\sqrt{3}\hat{k}}}}}$

To Find :

▪ Magnitude of the given vector quantity.

SoluTion :

If Ax, Ay, Az are the rectangular components of vector A and î, j, k are are the unit vectors along X-, Y- and Z-axis respectively, then

$</p><p>\dag\bf\:\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}† </p><p>$

Magnitude of vector :

$\begin{gathered}\dashrightarrow\bf\:A=\sqrt{(A_x)^2+(A_y)^2+(A_z)^2}\\ \\ \dashrightarrow\sf\:A=\sqrt{(-3)^2+(2)^2+(\sqrt{3})^2}\\ \\ \dashrightarrow\sf\:A=\sqrt{9+4+3}\\ \\ \dashrightarrow\sf\:A=\sqrt{16}\\ \\ \dashrightarrow\underline{\boxed{\bf{\green{A=4\:unit}}}}\:\orange{\bigstar}\end{gathered} </p><p>$

$\huge \sf \pink{Hence \: Done !!}$