If three consecutive vertices of a parallelogram are $A(4,3,5)$,$B(0,6,0)$,$C(-8,1,4)$ and $D$ is the fourth vertex,then the angle between $AC$ and $BD$ is

If $a \cdot i = 4$,then $(a \times j) \cdot (2j - 3k) = $

Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c}+\vec{d}|=\sqrt{29}$ and $\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$. If $\lambda_1, \lambda_2$ $(\lambda_1 > \lambda_2)$ are the possible values of $(\vec{c}+\vec{d}) \cdot (-7\hat{i}+2\hat{j}+3\hat{k})$, then the equation $K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+(3K+\frac{\lambda_{2}}{2})y^{2}-8x+12y+\lambda_{2}=0$ represents a circle, for $K$ equal to:

If $\vec{a}$ is a nonzero vector such that its projections on the vectors $2 \hat{i}-\hat{j}+2 \hat{k}$,$\hat{i}+2 \hat{j}-2 \hat{k}$,and $\hat{k}$ are equal,then a unit vector along $\vec{a}$ is:

$A$ particle acted on by constant forces $4i + j - 3k$ and $3i + j - k$ is displaced from the point $i + 2j + 3k$ to the point $5i + 4j + k$. The total work done by the force is ............... $unit$.

Forces $3i + 2j + 5k$ and $2i + j - 3k$ are acting on a particle and displace it from the point $2i - j - 3k$ to the point $4i - 3j + 7k$. The work done by the forces is ............... $unit$.