Let O be the origin.
\(\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}, \overrightarrow{O C}=\vec{c}\)
Since D and E are the mid points of AB and AC
\( \therefore \overrightarrow{O D} =\frac{\overrightarrow{O A}+\overrightarrow{O B}}{2}=\frac{\vec{a}+\vec{b}}{2} ; \overrightarrow{O E}=\frac{\overrightarrow{O A}+\overrightarrow{O C}}{2}=\frac{\vec{a}+\vec{c}}{2} \)
\(\text { LHS } =\overrightarrow{B E}+\overrightarrow{D C}=\overrightarrow{O E}-\overrightarrow{O B}+\overrightarrow{O C}-\overrightarrow{O D} \)
\( =\frac{\vec{a}+\vec{c}}{2}-\vec{b}+\vec{c}-\frac{(\vec{a}+\vec{b})}{2} \)
\( =\frac{\vec{a}+\vec{c}-2 \vec{b}+2 \vec{c}-\vec{a}-\vec{b}}{2} \)
\( =\frac{3 \vec{c}-3 \vec{b}}{2}=\frac{3(\vec{c}-\vec{b})}{2}=\frac{3(\overrightarrow{O C}-\overrightarrow{O B})}{2} \)
\( =\frac{3}{2} \overrightarrow{B C}=R H S\)
Hence proved.

Given \(\vec{a}=\hat{i}+2\hat{j}+3\hat{k}\) and \(\vec{b}=2\hat{i}+3\hat{j}-5\hat{k}\)
\(\therefore \left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & 3 \\ 2 & 3 & -5 \end{matrix} \right| =\hat { i } (-10-9)-\hat { j(-5-6)+\hat { k } (3-4)=-19\hat { i } +11\hat { j } -\hat { k } } \)
Now, \(\vec { a } .(\vec { a } \times \vec { b } )=(\hat { i } +2\hat { j } +3\hat { k } ).(-19\hat { i } +11\hat { j } -\hat { k } )=-19\hat { i } +11\hat { j } -\hat { k } \)
=-19+22-3=-22+22=0
This shows \(\vec{a}\) and \(\vec{a} \times \vec{b}\) are perpendicular to each other.

(i) Time - period - scalar
(ii) distance - scalar
(iii) force - vector
(iv) velocity - vector
(v) work done - scalar

Let \(l={4\over3} , m=o,n={3\over4}\)
\(\therefore l^2+m^2+n^2=({4\over3})^2+0^2+({3\over4})^2={16\over 9}+{9\over 16}={256+9\over 16\times9}\)\(={265\over144}\neq 1\)
Hence, the given ratios are not the direction cosines of any vector.

Let A and B be the points (2, 3, 1) and (3,-1, 2).
The direction cosines of\(\overrightarrow{AB},\) are \({1\over \sqrt{18}},{-4\over \sqrt{18}},{1\over \sqrt{18}}\).
But any point can be taken as first point.
Hence we have another set of direction cosines with opposite direction.
Thus, we have another set of direction ratio \({-1\over \sqrt{18}},{4\over \sqrt{18}},{-1\over \sqrt{18}}.\)

We know that a unit vector along the direction of the vector \(\overrightarrow{a}\) is given by \({\overrightarrow{a}\over|\overrightarrow{a}|}\). 
So a unit vector along the direction of 5 \(\hat{i}\) - 3 \(\hat{j}\) + 4 \(\hat{k}\) is given by \({5\hat{i}-3\hat{j}+4\hat{k}\over |5\hat{i}-3\hat{j}+4\hat{k}|}={5\hat{i}-3\hat{j}+4\hat{k}\over\sqrt{5^2+3^2+4^2}}={5\hat{i}-3\hat{j}+4\hat{k}\over \sqrt{50}}\).

\(\overrightarrow{a}\times \overrightarrow{b}=\)\(\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} =\hat{i}(1+4)-\hat{j}(3-4)+\hat{k}(-3-1)=5\hat{i}+\hat{j}-4\hat{k}\)
\(|\overrightarrow{a}\times \overrightarrow{b}|=|5\hat{i}+​​\hat{j}-4\hat{k}|=\sqrt{42}\)
Area of the parallelogram is \(\sqrt{42}\) sq.units.

Let   \(2\hat{i}+3\hat{j}-6\hat{k}\) and \(6\hat{i}-3\hat{j}+2\hat{k}\)
Let \(\theta \) be the angle between the given vectors.
\(\overrightarrow{a}.\overrightarrow{b}=(2\hat{i}+3\hat{j}-6\hat{k}).(6\hat{i}-3\hat{j}+2\hat{k})\)
= 12 - 9 - 12 = -9
\(|\overrightarrow{a}|=\sqrt{2^2+3^2+(-6)^2}=\sqrt{4+9+36}=\sqrt{49}=7\)
and \(|\overrightarrow{b}|=\sqrt{6^2+(-3)^2+2^2}=\sqrt{36+9+4}=\sqrt{49}=7\)
\(\therefore cos \theta ={\overrightarrow{a}.\overrightarrow{b}\over |\overrightarrow{a}|.|\overrightarrow{b}|}={-9\over 7(7)}={-9\over 49}\)
\(\Rightarrow \theta =cos^{-1}({-9\over 49})\)

Given  | \(\overrightarrow{a}\) | = 10, | \(\overrightarrow{b}\) | = 15 and \(\overrightarrow{a}\).\(\overrightarrow{b}\) = 75 \(\sqrt{2}\)
Let \(\theta\) be the angle between the vector \(\overrightarrow{a}\) and \(\overrightarrow{b}\).

\(\theta ={\pi\over 4}.\)

The given vector is 5\(\hat{i}\) - 3\(\hat{j}\) - 48\(\hat{k}\)
The direction ratios are 5, -3, -48.
r = \(\sqrt{x^2+y^2+z^2}=\sqrt{5^2+(-3)^2+(-48)^2}\)
\(=\sqrt{25+9+2304}=\sqrt{2338}\)
Hence, the direction cosines are \({5\over \sqrt{2338}},{-3\over \sqrt{2338}},{-48\over \sqrt{2338}}\)