Find angle between $\vec A = 3\hat i - \hat j + 4\hat k$ and $Z-$ axis
${\tan ^{ - 1}}\,\left( {\frac{{\sqrt {22} }}{4}} \right)$
${\tan ^{ - 1}}\,\left( {\frac{{\sqrt {10} }}{4}} \right)$
${\sin ^{ - 1}}\,\left( {\frac{{\sqrt {10} }}{4}} \right)$
${\sin ^{ - 1}}\,\left( {\frac{4}{{\sqrt {26} }}} \right)$
Force $F$ applied on a body is written as $F =(\hat{ n } \cdot F ) \hat{ n }+ G$, where $\hat{ n }$ is a unit vector. The vector $G$ is equal to
If $|\vec A \times \vec B| = \sqrt 3 \vec A.\vec B,$ then the value of$|\vec A + \vec B|$ is
A vector has magnitude same as that of $\overrightarrow{\mathrm{A}}-=3 \hat{\mathrm{j}}+4 \hat{\mathrm{j}}$ and is parallel to $\overrightarrow{\mathrm{B}}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}$. The $\mathrm{x}$ and $y$ components of this vector in first quadrant are $\mathrm{x}$ and $3$ respectively where $X$=_____.
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. . . . . ..