If left| {vec A } right|, = ,2 and left| {vec B } right|, = ,4 then match relation in Column -

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3-1.Vectors

medium

If $\left| {\vec A } \right|\, = \,2$ and $\left| {\vec B } \right|\, = \,4$ then match the relation in Column $-I$ with the angle $\theta $ between $\vec A$ and $\vec B$ in Column $-II$.

Column $-I$ Column $-II$

$(a)$ $\vec A \,.\,\,\vec B \, = \,\,0$ $(i)$ $\theta = \,{0^o}$

$(b)$ $\vec A \,.\,\,\vec B \, = \,\,+8$ $(ii)$ $\theta = \,{90^o}$

$(c)$ $\vec A \,.\,\,\vec B \, = \,\,4$ $(iii)$ $\theta = \,{180^o}$

$(d)$ $\vec A \,.\,\,\vec B \, = \,\,-8$ $(iv)$ $\theta = \,{60^o}$

Option A

Option B

Option C

Option D

Solution

Given $|\mathrm{A}|=2$ and $|\mathrm{B}|=4$

$(a)$ $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB} \cos \theta=0$

$\therefore \quad 2 \times 4 \cos \theta=0$

$\therefore \quad \cos \theta=0=\cos 90^{\circ}$

$\therefore \quad \theta=90^{\circ}$

$\therefore$ Option $(a)$ matches with option $(ii)$.

$(b)$ $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB} \cos \theta=8$

$\therefore 2 \times 4 \cos \theta=8$

$\therefore \quad \cos \theta=1=\cos 0^{\circ}$

$\therefore \quad \theta=0^{\circ}$

$\therefore$ Option $(b)$ matches with option $(i)$.

$(c)$ $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB} \cos \theta=4$

$\therefore \quad 2 \times 4 \cos \theta=4$ $\therefore \quad \cos \theta=\frac{1}{2}=\cos 60^{\circ}$ $\therefore \quad \theta=60^{\circ}$

$\therefore$ Option $(c)$ matches with option $(iv)$.

$(d)$ $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB} \cos \theta=-8$

$\therefore \quad 2 \times 4 \cos \theta=-8$

$\therefore \quad \cos \theta=-1=\cos 180^{\circ}$

$\therefore \quad \cos \theta=180^{\circ}$

Option $(d)$ matches with option $(iii)$.

Standard 11

Physics

Vector $A$ is pointing eastwards and vector $B$ northwards. Then, match the following two columns.

Colum $I$ Colum $II$

$(A)$ $(A+B)$ $(p)$ North-east

$(B)$ $(A-B)$ $(q)$ Vertically upwards

$(C)$ $(A \times B)$ $(r)$ Vertically downwards

$(D)$ $(A \times B) \times(A \times B)$ $(s)$ None

medium

The value of $\hat{ i } \times(\hat{ i } \times a )+\hat{ j } \times(\hat{ j } \times a )+\hat{ k } \times(\hat{ k } \times a )$ is

medium

The angle between vectors $(\vec{M} \times \vec{N})$ and $(\bar{N} \times \vec{M})$ is …………….

easy

For three vectors $\vec{A}=(-x \hat{i}-6 \hat{j}-2 \hat{k})$, $\vec{B}=(-\hat{i}+4 \hat{j}+3 \hat{k})$ and $\vec{C}=(-8 \hat{i}-\hat{j}+3 \hat{k})$, if $\overrightarrow{\mathrm{A}} \cdot(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{C}})=0$, them value of $\mathrm{x}$ is. . . . . ..

hard

(JEE MAIN-2024)

Find the scalar and vector products of two vectors. $a =(3 \hat{ i }-4 \hat{ j }+5 \hat{ k })$ and $b =(- 2 \hat{ i }+\hat{ j }- 3 \hat { k } )$

medium

Column $-I$	Column $-II$
$(a)$ $\vec A \,.\,\,\vec B \, = \,\,0$	$(i)$ $\theta = \,{0^o}$
$(b)$ $\vec A \,.\,\,\vec B \, = \,\,+8$	$(ii)$ $\theta = \,{90^o}$
$(c)$ $\vec A \,.\,\,\vec B \, = \,\,4$	$(iii)$ $\theta = \,{180^o}$
$(d)$ $\vec A \,.\,\,\vec B \, = \,\,-8$	$(iv)$ $\theta = \,{60^o}$

Colum $I$	Colum $II$
$(A)$ $(A+B)$	$(p)$ North-east
$(B)$ $(A-B)$	$(q)$ Vertically upwards
$(C)$ $(A \times B)$	$(r)$ Vertically downwards
$(D)$ $(A \times B) \times(A \times B)$	$(s)$ None

Solution

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Vector $A$ is pointing eastwards and vector $B$ northwards. Then, match the following two columns.

Colum $I$ Colum $II$

$(A)$ $(A+B)$ $(p)$ North-east

$(B)$ $(A-B)$ $(q)$ Vertically upwards

$(C)$ $(A \times B)$ $(r)$ Vertically downwards

$(D)$ $(A \times B) \times(A \times B)$ $(s)$ None