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
| 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 |
- A$( A \rightarrow p , s , B \rightarrow s , C \rightarrow q , D \rightarrow s )$
- B$( A \rightarrow p , s , B \rightarrow r , C \rightarrow q , D \rightarrow s )$
- C$( A \rightarrow p , s , B \rightarrow q , C \rightarrow r , D \rightarrow s )$
- D$( A \rightarrow p , B \rightarrow s , C \rightarrow q , D \rightarrow s )$
Similar Questions
If $\overrightarrow A \times \overrightarrow B = \overrightarrow C ,$then which of the following statements is wrong
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. . . . . ..
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. . . . . ..
- [JEE MAIN 2024]
$\hat i.\left( {\hat j \times \,\,\hat k} \right) + \;\,\hat j\,.\,\left( {\hat k \times \hat i} \right) + \hat k.\left( {\hat i \times \hat j} \right)=$
$\hat i.\left( {\hat j \times \,\,\hat k} \right) + \;\,\hat j\,.\,\left( {\hat k \times \hat i} \right) + \hat k.\left( {\hat i \times \hat j} \right)=$
If $F _1$ and $F _2$ are two vectors of equal magnitudes $F$ such that $\left| F _1 \cdot F _2\right|=\left| F _1 \times F _2\right|$, then $\left| F _1+ F _2\right|$ equals to
A vector ${\overrightarrow F _1}$is along the positive $X-$axis. If its vector product with another vector ${\overrightarrow F _2}$ is zero then ${\overrightarrow F _2}$ could be
A vector ${\overrightarrow F _1}$is along the positive $X-$axis. If its vector product with another vector ${\overrightarrow F _2}$ is zero then ${\overrightarrow F _2}$ could be