Find the unit vector perpendicular to the plane containing the points $(1, -1, 2), (2, 0, -1),$ and $(0, 2, 1)$.

The vector $\vec{a}=-\hat{i}+2 \hat{j}+\hat{k}$ is rotated through a right angle,passing through the $y$-axis in its way and the resulting vector is $\vec{b}$. Then the projection of $3 \vec{a}+\sqrt{2} \vec{b}$ on $\vec{c}=5 \hat{i}+4 \hat{j}+3 \hat{k}$ is

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vertices of a triangle,show that $\frac{1}{2}[\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}]$ gives the vector area of the triangle. Hence,deduce the condition that the three points $\vec{a}, \vec{b}, \vec{c}$ are collinear. Also,find the unit vector normal to the plane of the triangle.

Let $x$ and $y$ be real numbers. If $\vec{a}=(\sin x) \hat{i}+(\sin y) \hat{j}$ and $\vec{b}=(\cos x) \hat{i}+(\cos y) \hat{j}$,then $|\vec{a} \times \vec{b}|$ is

Let $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$ be two given lines. Then the unit vector perpendicular to $L_1$ and $L_2$ is

If $a, b, c, d$ are coplanar vectors,then $(a \times b) \times (c \times d)$ is equal to