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 $\overrightarrow{a}=-\hat{i}+\hat{j}+2 \hat{k}$, $\overrightarrow{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{c}=-2 \hat{i}+\hat{j}+3 \hat{k}$, then the angle between $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$ is

If $3\vec{a} - 5\vec{b}$ and $2\vec{a} + \vec{b}$ are perpendicular to each other,and $\vec{a} + 4\vec{b}$ and $-\vec{a} + \vec{b}$ are also perpendicular to each other,and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,then find $\cos \theta$.

If $A, B, C$ and $D$ are four points in the plane such that $|AB|^2+|CD|^2=|BC|^2+|DA|^2=100$,then $AC \cdot BD=$

Let $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ with $a_{i} > 0$ for $i = 1, 2, 3$ be a vector that makes equal angles with the coordinate axes $OX$,$OY$,and $OZ$. Also,let the projection of $\vec{a}$ on the vector $3 \hat{i} + 4 \hat{j}$ be $7$. Let $\vec{b}$ be a vector obtained by rotating $\vec{a}$ by $90^{\circ}$. If $\vec{a}$,$\vec{b}$,and the $x$-axis are coplanar,then the projection of vector $\vec{b}$ on $3 \hat{i} + 4 \hat{j}$ is equal to

Let $\vec{a}$ and $\vec{b}$ be vectors of the same magnitude such that $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$. Then $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$ is: