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 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
If $\overrightarrow{ P }=3 \hat{ i }+\sqrt{3} \hat{ j }+2 \hat{ k }$ and $\overrightarrow{ Q }=4 \hat{ i }+\sqrt{3} \hat{ j }+2.5 \hat{ k }$ then, The unit vector in the direction of $\overrightarrow{ P } \times \overrightarrow{ Q }$ is $\frac{1}{x}(\sqrt{3} \hat{i}+\hat{j}-2 \sqrt{3} \hat{k})$. The value of $x$ is
Two forces ${\vec F_1} = 5\hat i + 10\hat j - 20\hat k$ and ${\vec F_2} = 10\hat i - 5\hat j - 15\hat k$ act on a single point. The angle between ${\vec F_1}$ and ${\vec F_2}$ is nearly ....... $^o$
Projection of vector $\vec A$ on $\vec B$ is
Find the angle between two vectors with the help of scalar product.