JEE Main Motion in Plane PYQ

21. (JEE Main 2021 (Online) 22th July Evening Shift )

What will be the projection of vector $\vec{A} = \hat{i} + \hat{j} + \hat{k}$ on vector $\vec{B} = \hat{i} + \hat{j}$ ?

(A) $\sqrt{2} (\hat{i} + \hat{j} + \hat{k})$

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

(C) $\sqrt{2} (\hat{i} + \hat{j})$

(D) $2 (\hat{i} + \hat{j} + \hat{k})$

Correct answer is (B)

Projection = $\frac{\vec{A} . \vec{B}}{| \vec{B} |} (\hat{B})$

$= \frac{(\hat{i} + \hat{j} + \hat{k}) . (\hat{i} + \hat{j})}{\sqrt{2}} \frac{(\hat{i} + \hat{j})}{\sqrt{2}}$

$= \frac{2}{\sqrt{2}}$ $\times$ $\frac{(\hat{i} + \hat{j})}{\sqrt{2}}$

$= (\hat{i} + \hat{j})$

22. (JEE Main 2021 (Online) 20th July Evening Shift )

Two vectors $\vec{P}$ and $\vec{Q}$ have equal magnitudes. If the magnitude of $\vec{P} + \vec{Q}$ is n times the magnitude of $\vec{P} - \vec{Q}$ , then angle between $\vec{P}$ and $\vec{Q}$ is :

(A) $\sin^{- 1} (\frac{n - 1}{n + 1})$

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

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

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

Correct answer is (D)

$| \vec{P} | = | \vec{Q} | = x$ ..... (i)

$| \vec{P} + \vec{Q} | = n | \vec{P} - \vec{Q} |$

$P^{2} + Q^{2} + 2 P Q \cos θ = n^{2} (P^{2} + Q^{2} - 2 P Q \cos θ)$

Using (i) in above equation

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

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

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

If $\vec{A}$ and $\vec{B}$ are two vectors satisfying the relation $\vec{A}$ . $\vec{B}$ = $| \vec{A} \times \vec{B} |$ . Then the value of $| \vec{A} - \vec{B} |$ will be :

(A) $\sqrt{A^{2} + B^{2} + \sqrt{2} A B}$

(B) $\sqrt{A^{2} + B^{2}}$

(C) $\sqrt{A^{2} + B^{2} - \sqrt{2} A B}$

(D) $\sqrt{A^{2} + B^{2} + 2 A B}$

Correct answer is (C)

Given, $\vec{A}$ . $\vec{B}$ = $| \vec{A} \times \vec{B} |$ ..... (i)

Also, we know that

$\vec{A}$ . $\vec{B}$ = $| \vec{A} | | \vec{B} |$ cos $θ$ .... (ii)

and $\vec{A} \times \vec{B}$ = $| \vec{A} | | \vec{B} |$ sin $θ$ ..... (iii)

From Eqs. (i), (ii) and (iii), we get

$| \vec{A} | | \vec{B} |$ cos $θ$ = $| \vec{A} | | \vec{B} |$ sin $θ$

$\Rightarrow \cos θ = \sin θ \Rightarrow \frac{\sin θ}{\cos θ} = 1$

$\Rightarrow \tan θ = 1$

$\Rightarrow \tan θ = \tan 45^{\circ} \Rightarrow θ = 45^{\circ}$

$∴$ $| \vec{A} - \vec{B} | = \sqrt{A^{2} + B^{2} - 2 | \vec{A} | | \vec{B} | \cos θ}$

$= \sqrt{A^{2} + B^{2} - 2 | \vec{A} | | \vec{B} | \cos (45^{\circ})}$

$= \sqrt{A^{2} + B^{2} - \sqrt{2} A B}$

24. (JEE Main 2021 (Online) 25th February Morning Shift )

In an octagon ABCDEFGH of equal side, what is the sum of

$\vec{A B} + \vec{A C} + \vec{A D} + \vec{A E} + \vec{A F} + \vec{A G} + \vec{A H}$ ,

if, $\vec{A O} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$

JEE Main 2021 (Online) 25th February Morning Shift Physics - Vector Algebra Question 23 English

(A) $- 16 \hat{i} - 24 \hat{j} + 32 \hat{k}$

(B) $16 \hat{i} + 24 \hat{j} - 32 \hat{k}$

(C) $16 \hat{i} + 24 \hat{j} + 32 \hat{k}$

(D) $16 \hat{i} - 24 \hat{j} + 32 \hat{k}$

Correct answer is (B)

We know,

$∵$ $\vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} + \vec{O E} + \vec{O F} + \vec{O G} + \vec{O H} = \vec{0}$

By triangle law of vector addition, we can write

$\vec{A B} = \vec{A O} + \vec{O B}; \vec{A C} = \vec{A O} + \vec{O C}$

$\vec{A D} = \vec{A O} + \vec{O D}; \vec{A E} = \vec{A O} + \vec{O E}$

$\vec{A F} = \vec{A O} + \vec{O F}; \vec{A G} = \vec{A O} + \vec{O G}$

$\vec{A H} = \vec{A O} + \vec{O H}$

Now

$\vec{A B} + \vec{A C} + \vec{A D} + \vec{A E} + \vec{A F} + \vec{A G} + \vec{A H}$

$= (7 \vec{A O}) + \vec{O B} + \vec{O C} + \vec{O D} + \vec{O E} + \vec{O F} + \vec{O G} + \vec{O H}$

$= (7 \vec{A O}) + \vec{0} - \vec{O A}$

$= (7 \vec{A O}) + \vec{A O}$

$= 8 \vec{A O} = 8 (2 \hat{i} + 3 \hat{j} - 4 \hat{k})$

$= 16 \hat{i} + 24 \hat{j} - 32 \hat{k}$

25. (JEE Main 2021 (Online) 25th February Evening Shift )

If $\vec{P} \times \vec{Q} = \vec{Q} \times \vec{P}$ , the angle between $\vec{P}$ and $\vec{Q}$ is $θ$ (0 $^{\circ}$ < $θ$ < 360 $^{\circ}$ ). The value of ' $θ$ ' will be ___________ $^{\circ}$ .

Correct answer is (180)

Given, $\vec{P} \times \vec{Q} = \vec{Q} \times \vec{P}$

It is possible only when $\vec{P} \times \vec{Q} = \vec{Q} \times \vec{P}$ = 0

$\Rightarrow$ We know, $\vec{P} \times \vec{Q}$ = PQsin $θ$

Only if $\vec{P} = 0$

or $\vec{Q} = 0$

or sin $θ$ = 0 $\Rightarrow$ $θ$ = 0 or 180^o

The angle b/w $\vec{P}$ & $\vec{Q}$ is $θ$ (0 $^{\circ}$ < $θ$ < 360 $^{\circ}$ )

So, $θ$ = 180 $^{\circ}$