Vector Algebra and Calculus

If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are collinear vectors, therefore they lie on the same line. Hence they are parallel. Hence there cross products will be 0.
Therefore
$(\vec{a}\times\vec{b})=(\vec{c}\times\vec{b})=(\vec{a}\times\vec{c})=0$
And application of the section formula gives us
$\vec{c}=\dfrac{\lambda \vec{a}+\mu\vec{b}}{\lambda+\mu}$
Or
$\vec{c}(\lambda+\mu)=\lambda \vec{a}+\mu\vec{b}$

Since, $\vec{a}+2\vec{b}$ is collinear with $\vec{c}$.
$\therefore \vec{a}+2\vec{b}=x\vec{c}, x\epsilon R$ and $\vec{b}+3\vec{c}$ is collinear with $\vec{a}$.
$\therefore \vec{b}+3\vec{c}=y\vec{a}, y\epsilon R$
$\Rightarrow \vec{a}+2\vec{b}+6\vec{c}=(1+2y)\vec{a}$
Also, $\vec{a}+2\vec{b}+6\vec{c}=(x+6)\vec{c}$
$\therefore (x+6)\vec{c}=(1+2y)\vec{a}$
$\Rightarrow x+6=0$
and $1+2y=0$
$\Rightarrow x=-6$ and $y=-1/2$
$\therefore \vec{a}+2\vec{b}+6\vec{c}=\vec{0}$

Consider the given points $A(\bar{a}),B(\bar{b}),C(\bar{c})$.
These three points determine two vectors $\vec{AB}$ and $\vec{AC}$

We know that, "two vectors $a,b$ are collinear if and only if $\vec{a} \times \vec{b}=0$".

Therefore two vectors $\vec{AB}$ and $\vec{AC}$ are collinear if and only if $\vec{AB} \times \vec{AC}=0$

$\vec{AB}=\vec{OB}-\vec{OA}=\bar{b}-\bar{a}$ and $\vec{AC}=\vec{OC}-\vec{OA}=\bar{c}-\bar{a}$

We have, two vectors $\vec{AB}$ and $\vec{AC}$ are collinear if and only if $\vec{AB} \times \vec{AC}=0$

$ \Rightarrow$ two vectors $\vec{AB}$ and $\vec{AC}$ are collinear if and only if $(\bar{b}-\bar{a}) \times (\bar{c}-\bar{a})=0$

Since the three points determine two vectors $\vec{AB}$ and $\vec{AC}$, we conclude that

Three points $A(\bar{a}),B(\bar{b}),C(\bar{c})$ are collinear if and only if $(\bar{b}-\bar{a}) \times (\bar{c}-\bar{a})=0$.

The position vector of COM of the.three particles will be given by
$\vec{r} _{COM}\, =\, \displaystyle \frac {m _1\vec{r} _1\, +\, m _2\vec{r} _2\, +\, m _3\vec{r} _3}{m _1\, +\, m _2\, +\, m _3}$
Substituting the values, we get
$\vec{r} _{COM}\, =\, \displaystyle \frac {(1) (\hat{i}\, +\, 4\hat{j}\, +\, \hat{k})\, +\, (2) (\hat{i}\, +\, \hat{j}\, +\, \hat{k})\, +\, (3) (2\hat{i}\, -\, \hat{j}\, -\, 2\hat{k})}{1+2+3}\, =\, \displaystyle \frac {1}{2}\, (3\hat{i}\, +\, \hat{j}\, -\, \hat{k})\, m$.
Hence, the position vector of their center of mass is $\, \displaystyle \frac {1}{2}\, (3\hat{i}\, +\, \hat{j}\, -\, \hat{k})\, m$.

Using vector geometry, the points P, Q, R divide the sides in given ratios. The sum of vectors AP, BQ, CR relates to the median CS. The ratio is 1/5 based on the geometric properties of the segments.

Collinear vectors are scalar multiples of each other, meaning vector b can be written as k times a. The magnitude of b is given as 21, and the magnitude of a is sqrt(2^2 + 3^2 + 6^2) = sqrt(4 + 9 + 36) = sqrt(49) = 7. Since the magnitude of b must be 21, the scalar multiplier k must be plus or minus 3, making the vector plus or minus 3(2i + 3j + 6k).

A vector is a mathematical or physical quantity that possesses both magnitude (size) and direction. Scalars only have magnitude, while vectors require both to be fully defined.

A resultant is the vector sum of two or more vectors. It represents the single vector that has the same effect as the individual vectors combined.

A vector quantity is defined as a quantity that has both magnitude and direction. Examples include velocity, force, and displacement.

The orthogonality condition (dot product = 0) and the obtuse angle condition (dot product with z-axis < 0) constrain the value of alpha. Solving the equations leads to the specified interval.

Plane ${P} _{1}$ is parallel to $\overrightarrow{a}$ and $\overrightarrow{b}$.
The normal to ${P} _{1}$ is along $\overrightarrow{a}\times \overrightarrow{b}$.
Plane ${P} _{2}$ is parallel to $\overrightarrow{c}$ and $\overrightarrow{d}$.
The normal to ${P} _{2}$ is along $\overrightarrow{c}\times \overrightarrow{d}$.
$\overrightarrow{A}$ is along the line of intersection of planes ${P} _{1}$ and ${P} _{2}$.
$\therefore \overrightarrow{A}$ is along $\left(\overrightarrow{a}\times\overrightarrow{b}\right)\times\left(\overrightarrow{c}\times\overrightarrow{d}\right)$
$\overrightarrow{a}\times\overrightarrow{b}=\left|\begin{matrix} \hat{i} &\hat{j}  &\hat{k}  \ 0 & 2 & 3 \ 0 &4  &-3  \end{matrix}\right|$
$=\left(-6-12\right)\hat{i}-0.\hat{j}+0.\hat{k}$ on simplification
$=-18\hat{i}$
$\left(\overrightarrow{a}\times\overrightarrow{b}\right)\times\left(\overrightarrow{c}\times\overrightarrow{d}\right)$
$\overrightarrow{c}\times\overrightarrow{d}=\left|\begin{matrix} \hat{i} &\hat{j}  &\hat{k}  \ 0 & 1 & -1 \ 3 &3  &0 \end{matrix}\right|$
$=\left(0+3\right)\hat{i}-\left(0+3\right)\hat{j}+\left(0+3\right)\hat{k}$ on simplification
$=3\hat{i}-3\hat{j}-3\hat{k}$
$=3\left(\hat{i}-\hat{j}-\hat{k}\right)$
The angle between $\overrightarrow{A}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is  $\theta$
$\cos{\theta}=\dfrac{\overrightarrow{A}}{\left|\overrightarrow{A}\right|}.\dfrac{\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3\sqrt{2}}$
$\pm\dfrac{\left(\hat{j}-\hat{k}\right).\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3\sqrt{2}}$
$\pm\dfrac{1}{\sqrt{2}}$
and $\cos{\theta}=\pm\dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta=\dfrac{\pi}{4},\dfrac{3\pi}{4}$

Plane ${p} _{1}$ is parallel to $\overrightarrow{a}$ and $\overrightarrow{b}$ the normal to ${p} _{1}$ is along $\overrightarrow{a}\times \overrightarrow{b}$ 
Plane ${p} _{2}$ is parallel to $\overrightarrow{c}$ and $\overrightarrow{d}$ the normal to ${p} _{2}$ is along $\overrightarrow{c}\times \overrightarrow{d}$
$\overrightarrow{A}$ is along the line of intersection of planes ${p} _{1}$ and ${p} _{2}$ 
$\therefore \overrightarrow{A}$ is along $\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{c}\times \overrightarrow{d}\right)$
$\overrightarrow{a}\times \overrightarrow{b}=\left[\begin{matrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 2 &  3\ 0 & 4 & -3 \end{matrix}\right]$
$=\hat{i}\left(-6-12\right)-\hat{j}\left(0-0\right)+\hat{k}\left(0\right)$
$=-18\hat{i}$
$\overrightarrow{c}\times \overrightarrow{d}=\left|\begin{matrix} \hat{i} & \hat{j} & \hat{k} \ 0  & 1 & -1 \ 3 & 3 & 0 \end{matrix}\right|$
$=\hat{i}\left(0+3\right)-\hat{j}\left(0+3\right)+\hat{k}\left(0-3\right)$
$=3\hat{i}-3\hat{j}-3\hat{k}$
$=3\left(\hat{i}-\hat{j}-\hat{k}\right)$
$ \therefore \overrightarrow{A}$ is along $\hat{i}\times \left(\hat{i}-\hat{j}-\hat{k}\right)=\hat{j}-\hat{k}$
The angle between $\overrightarrow{A}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is $\theta$
$\cos{\theta}=\dfrac{\overrightarrow{A}}{\left|\overrightarrow{A}\right|}.\dfrac{\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3}$
$   =\pm \dfrac{\left(\hat{j}-\hat{k}\right)\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3\sqrt{2}}$
$=\pm\dfrac{\left(1+2\right)}{3\sqrt{2}} = \pm \dfrac{1}{\sqrt{2}}$
and $\cos{\theta}=\pm \dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta=\dfrac{\pi}{4},\dfrac{3\pi}{4}$

The plane ${p} _{1}$ contains the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ into normal is along $\overrightarrow{a}\times \overrightarrow{b}$
The normal to plane ${p} _{2}$ is along  $\overrightarrow{c}\times \overrightarrow{d}$.
$\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{c}\times \overrightarrow{d}\right)=0$
$\Rightarrow$ two normals are parallel
$\therefore$ the angle between the planes is zero

Let plane $P$, contains $a,b$ vector
$\vec{n} _{1}=\ \vec{a}\times \vec{b}$
Plane $P _{2}$ contain $\vec{c},\vec{d}$ vector
$\vec{n} _{2}=\vec{c}\times \vec{d}$
If $ P _{1}\perp P _{2}$ than $ n _{1}\perp\ n _{2}$
$(\vec{a}\times \vec{b}).(\vec{c}\times \vec{d})=0$