Vector Algebra and Calculus

For vectors to be parallel, their components must be proportional: 2/6 = 3/alpha = 4/beta. 1/3 = 3/alpha => alpha = 9. 1/3 = 4/beta => beta = 12.

The vector perpendicular to the $i$ & $j$ plane would be along the unit vector $k$.
Another vector in $i$ & $j$ plane can be perpendicular to $\vec{F}$
$(3\widehat{i} +4\widehat{j}) \perp (4\widehat{i} -3\widehat{j}) $
But,  from the options only $7\widehat{k} \perp (4\widehat{i} -3\widehat{j})$

A vector $\displaystyle \vec{A}$ is parallel to an another vector $\displaystyle \vec{B}$ if it can be written as
$\displaystyle \vec{A}= m\vec{B}$ where $m$ is a constant.
Here, $\displaystyle \vec{A}=\left ( \hat{i}-\hat{j}+2\hat{k} \right )=\frac{1}{3}\left ( 3\hat{i}-3\hat{j}+6\hat{k} \right )$
or $\displaystyle \vec{A}=\frac{1}{3}\vec{B}$
This implies that $\vec A || \displaystyle \vec{B}$ and magnitude of $\displaystyle \vec{A}$ is $\displaystyle \frac{1}{3}$ times the magnitude of $\displaystyle \vec{B}.$

$\vec{a}\times\vec{b}=\begin{pmatrix}x _1\hat {i}+y _1\hat {j}\end{pmatrix}\times\begin{pmatrix}x _2\hat {i}+y _2\hat {j}\end{pmatrix}$

$\;\;\;\;\;\;\;\;\;\;\;=x _1y _2\hat {k}-x _2y _1\hat {k}=\vec{0}\Rightarrow x _1y _2=x _2y _1$

The vector product of two vectors $\vec{A}$ and $\vec{B}$ is defined by $\vec{A} \times \vec{B} = \hat{n} |A| |B| \sin x $, where  $\hat{n}$ is the unit vector perpendicular to both A and B vectors and x is the angle between them.
Here in this question the vector product of $\vec{P1}$ and $\vec{P2}$ vectors is zero. This is only possible in two cases: 1) Any of the vectors is zero itself or 2) the $\sin$ of the angle between them is zero.
From the given options, the vector parallel to the given vector  $\hat{i}$ is  $-4\hat{i}$. 

$\displaystyle \displaystyle \overrightarrow A .  \overrightarrow B = 0 \Rightarrow  \overrightarrow A \bot  \overrightarrow B$
and, $\overrightarrow A .  \overrightarrow C = 0 \Rightarrow  \overrightarrow A \bot  \overrightarrow C$
Also, $\overrightarrow B \times \overrightarrow C $ is perpendicular to both $\vec { B } \ and  \ \vec { C }$
$Thus, \overrightarrow { A } ||\, (\overrightarrow { B } \times \overrightarrow { C } )$

Magnitude of a vector A can be given as $ \sqrt{\overrightarrow{A}.\overrightarrow{A}} $. 
Hence, statement A is wrong .
Also, if $ \overrightarrow{a} = m \overrightarrow{b}$, taking dot product with b vector on both sides,
$ \overrightarrow{a}.\overrightarrow{b} = m \  b^2 \Rightarrow m = \dfrac {\overrightarrow{a}.\overrightarrow{b}}{b^2} $
Hence , statement B is correct.

$\vec{\mathrm{A}}=5\hat{\mathrm{i}}+6\hat{\mathrm{j}}+3\hat{\mathrm{k}}$  and $\vec{\mathrm{B}}=6\hat{\mathrm{i}}-2\hat{\mathrm{j}}-6\hat{\mathrm{k}}$  is given. 

Since, the vector a is parallel to b, the corresponding coefficients of all the 3 components must bear the same ratio
i.e $ \dfrac{2}{1} = \dfrac{6n}{18} = \dfrac{m}{3} $
Or, $6n = 36, n = 6$
And $m = 6$

Vector C is perpendicular to both vectors A and B. Hence, it can be equal to their cross product.
Options A and B make vector C in the plane of vectors A  and B which is not possible.
In option D, this is not possible as vector B is not perpendicular to  vector A

Component of A along B = (A dot B / |B|^2) * B. A dot B = 2-1 = 1. |B|^2 = 1^2 + (-1)^2 = 2. Component = (1/2)(i - j). Perpendicular component = A - (component along B) = (2i + j) - (0.5i - 0.5j) = 1.5i + 1.5j = (3/2)(i + j).

Given,

In order to make vectors a + b + c + d = 0, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.

Simple example:- a=î,b=2î,c=–3î,d=0

(b) Correct
a + b + c + d = 0
a + c = – (b + d)
Taking modulus on both the sides, we get:
| a + c | = | –(b + d)| = | b + d |
Hence, the magnitude of (a + c) is the same as the magnitude of (b + d).

(c) Correct
a + b + c + d = 0
a = – (b + c + d)
Taking modulus both sides, we get:
| a | = | b + c + d |
| a |  ≤  | a | + | b | + | c |  …. (#)

Equation (#) shows that the magnitude of a is equal to or less than the sum of the magnitudes of b, c, and d.
Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of b, c, and d.

(d) Correct
For a + b + c + d = 0
a + (b + c) + d = 0
The resultant sum of the three vectors a, (b + c), and d can be zero only if (b + c) lie in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle.

If a and d are collinear, then it implies that the vector (b + c) is in the line of a and d. This implication holds only then the vector sum of all the vectors will be zero.

This is the explanation of correct solution.

Three points with position vectors a, b, and c are collinear if (b-a) is a scalar multiple of (c-a). Given a = i - 2j + 3k, b = 2i + 3j - 4k, and c = 0i - 7j + 10k: b-a = i + 5j - 7k; c-a = -i - 5j + 7k. Since c-a = -1(b-a), the vectors are collinear.