The value of $b$ such that the scalar product of the vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ with the unit vector parallel to the sum of the vectors $\vec{u} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{v} = b\hat{i} + 2\hat{j} + 3\hat{k}$ is $1$, is...

$A$ particle is acted upon by constant forces $4\hat{i} + \hat{j} - 3\hat{k}$ and $3\hat{i} + \hat{j} - \hat{k}$. The displacement of the particle from the point $\hat{i} + 2\hat{j} + 3\hat{k}$ to the point $5\hat{i} + 4\hat{j} + \hat{k}$ is given. Find the total work done by the forces in units.

If $\triangle ABC$ is a right-angled triangle in which $BC$ is the hypotenuse, and the position vectors of $B$ and $C$ are $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{c} = 5\hat{i} + \hat{j} - 3\hat{k}$ respectively, then the value of $\vec{AB} \cdot \vec{AC} + \vec{BA} \cdot \vec{BC} + \vec{CA} \cdot \vec{CB}$ is:

Find the distance between the lines $l_{1}$ and $l_{2}$ given by $\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$ and $\vec{r}=3 \hat{i}+3 \hat{j}-5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+6 \hat{k})$.

$AB=a$ and $AC=b$ are the sides of $\triangle ABC$. $P$ is a point on $AB$ and $Q$ is a point on $BC$ such that $\frac{AP}{PB}=\frac{1}{2}$ and $\frac{BQ}{QC}=\frac{1}{2}$. If the point of intersection of $AQ$ and $CP$ is $D$ and the area of $\triangle BCD$ is $7$ square units,then the area of the $\triangle ABC$ (in the same square units) is

The shortest distance between the lines $r = 3i + 5j + 7k + \lambda(i + 2j + k)$ and $r = -i - j - k + \mu(7i - 6j + k)$ is