JEE Main Motion in Plane PYQ

11. (JEE Main 2022 (Online) 25th June Morning Shift )

Which of the following relations is true for two unit vector $\hat{A}$ and $\hat{B}$ making an angle $θ$ to each other?

(A) $| \hat{A} + \hat{B} | = | \hat{A} - \hat{B} | \tan \frac{θ}{2}$

(B) $| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \tan \frac{θ}{2}$

(C) $| \hat{A} + \hat{B} | = | \hat{A} - \hat{B} | c o s \frac{θ}{2}$

(D) $| \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | \cos \frac{θ}{2}$

Correct answer is (B)

$∵$ $| \hat{A} - \hat{B} | = 2 \sin (\frac{θ}{2})$

and, $| \hat{A} + \hat{B} | = 2 \cos (\frac{θ}{2})$

$\Rightarrow \frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \tan (\frac{θ}{2})$

12. (JEE Main 2022 (Online) 28th July Morning Shift )

If the projection of $2 \hat{i} + 4 \hat{j} - 2 \hat{k}$ on $\hat{i} + 2 \hat{j} + α \hat{k}$ is zero. Then, the value of $α$ will be ___________.

Correct answer is (5)

$\vec{A} = 2 \hat{i} + 4 \hat{j} - 2 \hat{k}$

$\vec{B} = \hat{i} + 2 \hat{j} + α \hat{k}$

$\vec{A} . \vec{B} = 0$ , as $\vec{A}$ should be perpendicular to $\vec{B}$

$\Rightarrow 2 + 8 - 2 α = 0$

$α = 5$

13. (JEE Main 2022 (Online) 26th July Evening Shift )

If $\vec{A} = (2 \hat{i} + 3 \hat{j} - \hat{k}) m$ and $\vec{B} = (\hat{i} + 2 \hat{j} + 2 \hat{k}) m$ . The magnitude of component of vector $\vec{A}$ along vector $\vec{B}$ will be ____________ $m$ .

Correct answer is (2)

$\begin{aligned} \vec{A} \cdot \vec{B} = (2 \hat{i} + 3 \hat{j} - \hat{k}) \cdot (\hat{i} + 2 \hat{j} + 2 \hat{k}) \\ \Rightarrow \vec{A} \cdot \vec{B} = 2 \times 1 + 3 \times 2 - 1 \times 2 \\ \Rightarrow \vec{A} \cdot \vec{B} = 6 \end{aligned}$

And $| \vec{B} | = \sqrt{1^{2} + 2^{2} + 2^{2}} = \sqrt{9} = 3$

Magnitude of component of

$\vec{A}$ along $\vec{B} = \frac{\vec{A} \cdot \vec{B}}{| B |} = \frac{6}{3} = 2$

14. (JEE Main 2021 (Online) 31st August Evening Shift )

Statement I :

Two forces $(\vec{P} + \vec{Q})$ and $(\vec{P} - \vec{Q})$ where $\vec{P} ⊥ \vec{Q}$ , when act at an angle $θ$ ₁ to each other, the magnitude of their resultant is $\sqrt{3 (P^{2} + Q^{2})}$ , when they act at an angle $θ$ ₂, the magnitude of their resultant becomes $\sqrt{2 (P^{2} + Q^{2})}$ . This is possible only when $θ_{1} < θ_{2}$ .

Statement II :

In the situation given above.

$θ$ ₁ = 60 $^{\circ}$ and $θ$ ₂ = 90 $^{\circ}$

In the light of the above statements, choose the most appropriate answer from the options given below :-

(A) Statement I is false but Statement II is true

(B) Both Statement I and Statement II are true

(C) Statement I is true but Statement II is false

(D) Both Statement I and Statement II are false.

Correct answer is (B)

$\vec{A} = \vec{P} + \vec{Q}$

$\vec{B} = \vec{P} - \vec{Q}$

$\vec{P} ⊥ \vec{Q}$

$| \vec{A} | = | \vec{B} | = \sqrt{2 (P^{2} + Q^{2}) (1 + \cos θ)}$

For $| \vec{A} + \vec{B} | = \sqrt{3 (P^{2} + Q^{2})}$

$θ_{1} = 60^{\circ}$

For $| \vec{A} + \vec{B} | = \sqrt{2 (P^{2} + Q^{2})}$

$θ_{2} = 90^{\circ}$

15. (JEE Main 2021 (Online) 27th August Morning Shift )

The resultant of these forces $\vec{O P}, \vec{O Q}, \vec{O R}, \vec{O S}$ and $\vec{O T}$ is approximately .......... N.

[Take $\sqrt{3} = 1.7$ , $\sqrt{2} = 1.4$ Given $\hat{i}$ and $\hat{j}$ unit vectors along x, y axis]

JEE Main 2021 (Online) 27th August Morning Shift Physics - Vector Algebra Question 13 English

(A) $9.25 \hat{i} + 5 \hat{j}$

(B) $3 \hat{i} + 15 \hat{j}$

(C) $2.5 \hat{i} - 14.5 \hat{j}$

(D) $- 1.5 \hat{i} - 15.5 \hat{j}$

Correct answer is (A)

JEE Main 2021 (Online) 27th August Morning Shift Physics - Vector Algebra Question 13 English Explanation
$\vec{F_{x}} = (10 \times \frac{\sqrt{3}}{2} + 20 (\frac{1}{2}) + 20 (\frac{1}{\sqrt{2}}) - 15 (\frac{1}{\sqrt{2}}) - 15 (\frac{\sqrt{3}}{2})) \hat{i}$

$= 9.25 \hat{i}$

$\vec{F_{y}} = (15 (\frac{1}{2}) + 20 (\frac{\sqrt{3}}{2}) + 10 (\frac{1}{2}) - 15 (\frac{1}{\sqrt{2}}) - 20 (\frac{1}{\sqrt{2}})) \hat{j}$

$= 5 \hat{j}$