JEE Main Motion in Plane PYQ

16. (JEE Main 2021 (Online) 26th August Evening Shift )

The angle between vector $(\vec{A})$ and $(\vec{A} - \vec{B})$ is :

JEE Main 2021 (Online) 26th August Evening Shift Physics - Vector Algebra Question 14 English

(A) $\tan^{- 1} (\frac{- \frac{B}{2}}{A - B \frac{\sqrt{3}}{2}})$

(B) $\tan^{- 1} (\frac{A}{0.7 B})$

(C) $\tan^{- 1} (\frac{\sqrt{3} B}{2 A - B})$

(D) $\tan^{- 1} (\frac{B \cos θ}{A - B \sin θ})$

Correct answer is (C)

JEE Main 2021 (Online) 26th August Evening Shift Physics - Vector Algebra Question 14 English Explanation

Angle between $\vec{A}$ and $\vec{B}$ , $θ$ = 60 $^{\circ}$

Angle between $\vec{A}$ and - $\vec{B}$ , $θ$ = 120 $^{\circ}$

If angle between $\vec{A}$ and $\vec{A}$ $-$ $\vec{B}$ is $α$

then $\tan α =$ $\frac{| - \vec{B} | \sin θ}{\vec{A} + | - \vec{B} | \cos θ}$

= $\frac{B \sin 120^{\circ}}{A + B \cos 120^{\circ}}$

= $\frac{B \frac{\sqrt{3}}{2}}{A - \frac{B}{2}}$

$\Rightarrow$ $\tan α = \frac{\sqrt{3} B}{2 A - B}$

17. (JEE Main 2021 (Online) 26th August Morning Shift )

The magnitude of vectors $\vec{O A}$ , $\vec{O B}$ and $\vec{O C}$ in the given figure are equal. The direction of $\vec{O A}$ + $\vec{O B}$ $-$ $\vec{O C}$ with x-axis will be :

JEE Main 2021 (Online) 26th August Morning Shift Physics - Vector Algebra Question 15 English

(A) $\tan^{- 1} \frac{(1 - \sqrt{3} - \sqrt{2})}{(1 + \sqrt{3} + \sqrt{2})}$

(B) $\tan^{- 1} \frac{(\sqrt{3} - 1 + \sqrt{2})}{(1 + \sqrt{3} - \sqrt{2})}$

(C) $\tan^{- 1} \frac{(\sqrt{3} - 1 + \sqrt{2})}{(1 - \sqrt{3} + \sqrt{2})}$

(D) $\tan^{- 1} \frac{(1 + \sqrt{3} - \sqrt{2})}{(1 - \sqrt{3} - \sqrt{2})}$

Correct answer is (A)

JEE Main 2021 (Online) 26th August Morning Shift Physics - Vector Algebra Question 15 English Explanation
Let magnitude be equal to $λ$

$\vec{O A} = λ [\cos 30^{\circ} \hat{i} + \sin 30 \hat{j}] = λ [\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}]$

$\vec{O B} = λ [\cos 60^{\circ} \hat{i} - \sin 60 \hat{j}] = λ [\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}]$

$\vec{O C} = λ [\cos 45^{\circ} (- \hat{i}) + \sin 45 \hat{j}] = λ [- \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}]$

$∴$ $\vec{O A} + \vec{O B} - \vec{O C}$

$= λ [(\frac{\sqrt{3} + 1}{2} + \frac{1}{\sqrt{2}}) \hat{i} + (\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}) \hat{j}]$

$∴$ Angle with x-axis

$\tan^{- 1} [\frac{\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}}}] = \tan^{- 1} [\frac{\sqrt{2} - \sqrt{6} - 2}{\sqrt{6} + \sqrt{2} + 2}]$

$= \tan^{- 1} [\frac{1 - \sqrt{3} - \sqrt{2}}{\sqrt{3} + 1 + \sqrt{2}}]$

Hence option (a).

18. (JEE Main 2021 (Online) 27th July Morning Shift )

Assertion A : If A, B, C, D are four points on a semi-circular are with centre at 'O' such that $| \vec{A B} | = | \vec{B C} | = | \vec{C D} |$ , then $\vec{A B} + \vec{A C} + \vec{A D} = 4 \vec{A O} + \vec{O B} + \vec{O C}$

Reason R : Polygon law of vector addition yields $\vec{A B} + \vec{B C} + \vec{C D} + \vec{A D} = 2 \vec{A O}$

JEE Main 2021 (Online) 27th July Morning Shift Physics - Vector Algebra Question 16 English
In the light of the above statements, choose the most appropriate answer from the options given below :

(A) A is correct but R is not correct.

(B) A is not correct but R is correct.

(C) Both A and R are correct and R is the correct explanation of A.

(D) Both A and R are correct but R is not the correct explanation of A.

Correct answer is (D)

Polygon law is applicable in both but the equation given in the reason is not useful in explaining the assertion.

19. (JEE Main 2021 (Online) 25th July Evening Shift )

Two vectors $\vec{X}$ and $\vec{Y}$ have equal magnitude. The magnitude of ( $\vec{X}$ $-$ $\vec{Y}$ ) is n times the magnitude of ( $\vec{X}$ + $\vec{Y}$ ). The angle between $\vec{X}$ and $\vec{Y}$ is :

(A) $\cos^{- 1} (\frac{- n^{2} - 1}{n^{2} - 1})$

(B) $\cos^{- 1} (\frac{n^{2} - 1}{- n^{2} - 1})$

(C) $\cos^{- 1} (\frac{n^{2} + 1}{- n^{2} - 1})$

(D) $\cos^{- 1} (\frac{n^{2} + 1}{n^{2} - 1})$

Correct answer is (B)

Given X = Y

$\sqrt{X^{2} + Y^{2} - 2 \times Y \cos θ}$

$= n \sqrt{X^{2} + Y^{2} + 2 \times Y \cos θ}$

Square both sides

$2 X^{2} (1 - \cos θ) = n^{2} .2 X^{2} (1 + \cos θ)$

$1 - \cos θ = n^{2} + n^{2} \cos θ$

$\cos θ = \frac{1 - n^{2}}{1 + n^{2}}$

$θ = \cos^{- 1} [\frac{n^{2} - 1}{- n^{2} - 1}]$

20. (JEE Main 2021 (Online) 25th July Morning Shift )

Match List - I with List - II

JEE Main 2021 (Online) 25th July Morning Shift Physics - Vector Algebra Question 18 English
Choose the correct answer from the options given below :

(A) (a) $\to$ (iv), (b) $\to$ (i), (c) $\to$ (iii), (d) $\to$ (ii)

(B) (a) $\to$ (iv), (b) $\to$ (iii), (c) $\to$ (i), (d) $\to$ (ii)

(C) (a) $\to$ (iii), (b) $\to$ (ii), (c) $\to$ (iv), (d) $\to$ (i)

(D) (a) $\to$ (i), (b) $\to$ (iv), (c) $\to$ (ii), (d) $\to$ (iii)

Correct answer is (B)

(a) $\vec{C} = \vec{A} + \vec{B}$

Option (iv)

(b) $\vec{A} = \vec{B} + \vec{C} = \vec{C} + \vec{B}$

Option (iii)

(c) $\vec{B} = \vec{A} + \vec{C}$

Option (i)

(d) $\vec{A} + \vec{B} + \vec{C} = 0$

Option (ii)