A vector $\overrightarrow{ A }$ points vertically upward and $\overrightarrow{ B }$ points towards north. The vector product $\overrightarrow{ A } \times \overrightarrow{ B }$ is
- Anull vector
- Balong west
- Calong east
- Dverticaly downward
Similar Questions
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
Show that the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.
Show that the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.
$\overrightarrow A $ and $\overrightarrow B $ are two vectors given by $\overrightarrow A = 2\widehat i + 3\widehat j$ and $\overrightarrow B = \widehat i + \widehat j$. The magnitude of the component (projection) of $\overrightarrow A$ on $\overrightarrow B$ is
$\overrightarrow A $ and $\overrightarrow B $ are two vectors given by $\overrightarrow A = 2\widehat i + 3\widehat j$ and $\overrightarrow B = \widehat i + \widehat j$. The magnitude of the component (projection) of $\overrightarrow A$ on $\overrightarrow B$ is
Let $\left| {{{\vec A}_1}} \right| = 3,\,\left| {\vec A_2} \right| = 5$, and $\left| {{{\vec A}_1} + {{\vec A}_2}} \right| = 5$. The value of $\left( {2{{\vec A}_1} + 3{{\vec A}_2}} \right)\cdot \left( {3{{\vec A}_1} - 2{{\vec A}_2}} \right)$ is
Let $\left| {{{\vec A}_1}} \right| = 3,\,\left| {\vec A_2} \right| = 5$, and $\left| {{{\vec A}_1} + {{\vec A}_2}} \right| = 5$. The value of $\left( {2{{\vec A}_1} + 3{{\vec A}_2}} \right)\cdot \left( {3{{\vec A}_1} - 2{{\vec A}_2}} \right)$ is
- [JEE MAIN 2019]
If a vector $2\hat i + 3\hat j + 8\hat k$ is perpendicular to the vector $4\hat j - 4\hat i + \alpha \hat k$. Then the value of $\alpha $ is
If a vector $2\hat i + 3\hat j + 8\hat k$ is perpendicular to the vector $4\hat j - 4\hat i + \alpha \hat k$. Then the value of $\alpha $ is
- [AIPMT 2005]