Question

Please solve these questions of Physics i have to submit my assignment soon . (With working not the direct answer)
I will mark you as brainliest if you help me.​​

Answer 1

Solutions:

1). Given that -

• $\sf{\vec{r} = (3 \hat{i} + 4 \hat{j} + 7 \hat{k})}$

We've to find it's magnitude.

$\longrightarrow$ $\sf{Magnitude\:=\:\sqrt{3^{2} + 4^{2} + 7^{2}}}$

$\longrightarrow$ $\sf{Magnitude\:=\:\sqrt{9 + 16 + 49}}$

$\longrightarrow$ $\sf{Magnitude\:=\:\sqrt{74}\:metre}$

Therefore, magnitude of the given vector will be √74 m.

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2). Given vectors -

• $\sf{\vec{A} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}}$
• $\sf{\vec{B} = \hat{i} + 2 \hat{j} + 3 \hat{k}}$

We've to find the unit vector parallel to the resultant of given vectors.

Resultant addition of vectors, R = A + B

$\longrightarrow$ $\sf{R = (2 \hat{i} + 4 \hat{j} - 5 \hat{k}) + (\hat{i} + 2 \hat{j} + 3 \hat{k})}$

$\longrightarrow$ $\sf{R = 2 \hat{i} + 4 \hat{j} - 5 \hat{k} + \hat{i} + 2 \hat{j} + 3 \hat{k}}$

$\longrightarrow$ $\sf{R = 3 \hat{i} + 6 \hat{j} - 2 \hat{k}}$

Magnitude -

$\longrightarrow$ $\sf{Magnitude\:=\:\sqrt{3^{2} + 6^{2} + (-2)^{2}}}$

$\longrightarrow$ $\sf{Magnitude\:=\:\sqrt{9 + 36 + 4}}$

$\longrightarrow$ $\sf{Magnitude\:=\:\sqrt{49}}$

$\longrightarrow$ $\sf{Magnitude = 7\:units}$

Now,

Unit vector is the ratio of complete vector and Its magnitude.

$\large{\boxed{\sf{\blue{Unit\:vector\:=\:\dfrac{3 \hat{i} + 6 \hat{j} - 2 \hat{k}}{7}}}}}$

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3). Given vectors -

• $\sf{\vec{A} = \hat{i} + 2 \hat{j} + 3 \hat{k}}$
• $\sf{\vec{B} = 2 \hat{i} - \hat{j}}$

Magnitude of A -

$\longrightarrow$ $\sf{Magnitude\:of\:A\:=\:\sqrt{1^{2} + 2^{2} + 3^{2}}}$

$\longrightarrow$ $\sf{Magnitude\:of\:A\:=\:\sqrt{1 + 4 + 9}}$

$\longrightarrow$ $\sf{Magnitude\:of\:A\:=\:\sqrt{14}}$

Magnitude of B -

$\longrightarrow$ $\sf{Magnitude\:of\:B\:=\:\sqrt{2^{2} + (-1)^{2}}}$

$\longrightarrow$ $\sf{Magnitude\:of\:B\:=\:\sqrt{4 + 1}}$

$\longrightarrow$ $\sf{Magnitude\:of\:B\:=\:\sqrt{5}}$

We know that,

$\sf{\vec{A} . \vec{B} = AB \cos \theta}$

Subsituting the values,

$\longrightarrow$ $\sf{(\hat{i} + 2 \hat{j} + 3 \hat{k}).(2 \hat{i} - \hat{j}) = \sqrt{14} \sqrt{15} \cos \theta}$

$\longrightarrow$ $\sf{\hat{i} (2 \hat{i} - \hat{j}) + 2 \hat{j} (2 \hat{i} - \hat{j}) + 3 \hat{k} (2 \hat{i} - \hat{j}) = \sqrt{14} \sqrt{15} \cos \theta}$

$\longrightarrow$ $\sf{2 (\hat{i} . \hat{i}) - (\hat{i} . \hat{j}) + 4 (\hat{i} . \hat{j}) - 2 (\hat{j} . \hat{j}) + 6 (\hat{i} . \hat{k}) - 3 (\hat{k} . \hat{j}) = \sqrt{14} \sqrt{15} \cos \theta}$

$\longrightarrow$ $\sf{2(1) - (0) + 4(0) - 2(1) + 6(0) - 3(0) = \sqrt{14} \sqrt{15} \cos \theta}$

$\longrightarrow$ $\sf{2 - 2 = \sqrt{14} \sqrt{15} \cos \theta}$

$\longrightarrow$ $\sf{\cos \theta = 0}$

$\longrightarrow$ $\sf{\cos 90^{\circ} = 0}$

$\longrightarrow$ $\sf{\theta = 90^{\circ}}$

Therefore, given vectors are perpendicular to each other.

Hence proved!

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Points to remember -

• $\sf{\hat{i} . \hat{i} = 1}$
• $\sf{\hat{j} . \hat{j} = 1}$
• $\sf{\hat{k} . \hat{k} = 1}$
• $\sf{\hat{i} . \hat{j} = 0}$
• $\sf{\hat{j} . \hat{k} = 0}$
• $\sf{\hat{k} . \hat{i} = 0}$

Answer 2

AnswEr :

4) Given,

$\sf \vec{f} = \hat{i} + 2 \hat{j} - \hat{k} \\ \sf \vec{s} = 4 \hat{i} - \hat{j} + 7 \hat{k}$

Workdone is the cross product of force and displacement vector.

$\sf W = \vec{f}. \vec{s}$

We know that,

$\sf \vec{a}. \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$

Therefore,

$\longrightarrow \sf \:W = 4 - 2 - 7 \\ \\ \longrightarrow \sf \: W = - 5 \: J$

5) Given,

$\sf \vec{A} = 2 \hat{I} - 3\hat{j} + 4 \hat{k} \\ \sf \vec{B} = -6 \hat{i} + 9\hat{j} -12 \hat{k}$

Here,

$\sf \vec{B} = - 3 \vec{A}$

Thus, the two vectors are parallel to eachother.

6) Refer to the Attachment.