If the area of a parallelogram whose diagonals are represented by vectors $\vec{d_1} = 3 \hat{i} + \lambda \hat{j} + 2 \hat{k}$ and $\vec{d_2} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ is $\frac{\sqrt{117}}{2}$ sq. units,then $\lambda=$

If $\overline{a}$ and $\overline{b}$ are two unit vectors such that $5 \overline{a} + 4 \overline{b}$ and $\overline{a} - 2 \overline{b}$ are perpendicular to each other,then the angle between $\overline{a}$ and $\overline{b}$ is

If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b},$ then $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$ when $\theta$ is equal to

Let $ABC$ be a triangle. Let a point $P$ divide $AB$ in the ratio $1:2$ internally and a point $Q$ divide $BC$ in the ratio $1:2$ internally. Let $D$ be the point of intersection of $AQ$ and $CP$. If the area of the triangle $ABC$ is $k$ square units,then the area of the triangle $BCD$ in square units is:

If $\theta$ is the angle between the vectors $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,then:

Let $a=\hat{i}+2 \hat{j}-2 \hat{k}$ and $b=2 \hat{i}-\hat{j}-2 \hat{k}$. If the orthogonal projection vector of $a$ on $b$ is $x$ and the orthogonal projection vector of $b$ on $a$ is $y$, then $|x-y|=$