Three particles ${P}, {Q}$ and ${R}$ are moving along the vectors ${A}=\hat{{i}}+\hat{{j}}, {B}=\hat{{j}}+\hat{{k}}$ and ${C}=-\hat{{i}}+\hat{{j}}$ respectively. They strike on a point and start to move in different directions. Now particle $P$ is moving normal to the plane which contains vector $\vec{A}$ and $\vec{B} .$ Similarly particle $Q$ is moving normal to the plane which contains vector $\vec{A}$ and $\vec{C} .$ The angle between the direction of motion of $P$ and $Q$ is $\cos ^{-1}\left(\frac{1}{\sqrt{x}}\right)$. Then the value of $x$ is ...... .
$11$
$47$
$5$
$3$
The area of the triangle formed by $2\hat i + \hat j - \hat k$ and $\hat i + \hat j + \hat k$ is
A vector $\overrightarrow A $ points vertically upward and $\overrightarrow B $points towards north. The vector product $\overrightarrow A \times \overrightarrow B $ is
If $\vec A = 2\hat i + \hat j - \hat k,\,\vec B = \hat i + 2\hat j + 3\hat k$ and $\vec C = 6\hat i - 2j - 6\hat k$ then the angle between $(\vec A + \vec B)$ and $\vec C$ wil be ....... $^o$