Class 12: Algebra of Vectors

$\displaystyle \textbf{Question 1: }~\text{Find the magnitude of the vector }\vec{a}=2\hat{i}+3\hat{j}-6\hat{k}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: }\overrightarrow{a}=2\widehat{i}+3\widehat{j}-6\widehat{k}$
$\displaystyle \text{Magnitude of the vector }|\overrightarrow{a}|=\sqrt{2^{2}+3^{2}+(-6)^{2}}=\sqrt{4+9+36}=\sqrt{49}=7$

$\displaystyle \textbf{Question 2: }~\text{Find the unit vector in the direction of }3\hat{i}+4\hat{j}-12\hat{k}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=3\widehat{i}+4\widehat{j}-12\widehat{k}$
$\displaystyle \text{Then,}$
$\displaystyle |\overrightarrow{a}|=\sqrt{3^{2}+4^{2}+(-12)^{2}}=\sqrt{9+16+144}=\sqrt{169}=13$
$\displaystyle \text{So, a unit vector in the direction of }\overrightarrow{a}\text{ is given by}$
$\displaystyle \hat{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{1}{13}(3\widehat{i}+4\widehat{j}-12\widehat{k})=\frac{3}{13}\widehat{i}+\frac{4}{13}\widehat{j}-\frac{12}{13}\widehat{k}$

$\displaystyle \textbf{Question 3: }~\text{Find a unit vector in the direction of the resultant of the vectors }\hat{i}-\hat{j}+3\hat{k},\ 2\hat{i}+\hat{j}-2\hat{k}\text{ and }\hat{i}+2\hat{j}-2\hat{k}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: }\overrightarrow{a}=\widehat{i}-\widehat{j}+3\widehat{k},\;\overrightarrow{b}=2\widehat{i}+\widehat{j}-2\widehat{k}\text{ and }\overrightarrow{c}=\widehat{i}+2\widehat{j}-2\widehat{k}.$
$\displaystyle \text{Then, Resultant of the vectors }=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$
$\displaystyle =\widehat{i}-\widehat{j}+3\widehat{k}+2\widehat{i}+\widehat{j}-2\widehat{k}+\widehat{i}+2\widehat{j}-2\widehat{k}$
$\displaystyle =4\widehat{i}+2\widehat{j}-\widehat{k}$
$\displaystyle \text{So,}$
$\displaystyle |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt{4^{2}+2^{2}+(-1)^{2}}=\sqrt{16+4+1}=\sqrt{21}$
$\displaystyle \text{Unit vector in the direction of the resultant vector }=$
$\displaystyle \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|}=\frac{1}{\sqrt{21}}(4\widehat{i}+2\widehat{j}-\widehat{k})$

$\displaystyle \textbf{Question 4: }~\text{The adjacent sides of a parallelogram are represented by the vectors }\vec{a}=\hat{i}+\hat{j}-\hat{k}\text{ and }\vec{b}=-2\hat{i}+\hat{j}+2\hat{k}. \text{ Find unit vectors parallel to the diagonals of the parallelogram.}$
$\displaystyle \text{Answer:}$
$\displaystyle \overrightarrow{a}=\widehat{i}+\widehat{j}-\widehat{k}\text{ and }\overrightarrow{b}=-2\widehat{i}+\widehat{j}+2\widehat{k}.$
$\displaystyle \overrightarrow{AC}=\overrightarrow{a}+\overrightarrow{b}=-\widehat{i}+2\widehat{j}+\widehat{k}$
$\displaystyle \text{Let the unit vector along the diagonals }AC\text{ and }BD \\ \text{ of the parallelogram be }\widehat{AC}\text{ and }\widehat{BD}.$
$\displaystyle \widehat{AC}=\frac{-\widehat{i}+2\widehat{j}+\widehat{k}}{\sqrt{(-1)^{2}+2^{2}+1^{2}}}=\frac{-\widehat{i}+2\widehat{j}+\widehat{k}}{\sqrt{6}}$
$\displaystyle \text{Similarly, }\overrightarrow{BD}=\overrightarrow{b}-\overrightarrow{a}=-3\widehat{i}+3\widehat{k}$
$\displaystyle \widehat{BD}=\frac{-3\widehat{i}+3\widehat{k}}{\sqrt{(-3)^{2}+0^{2}+3^{2}}}=\frac{-3\widehat{i}+3\widehat{k}}{3\sqrt{2}}=\frac{-\widehat{i}+\widehat{k}}{\sqrt{2}}$

$\displaystyle \textbf{Question 5: }~\text{If }\vec{a}=3\hat{i}-\hat{j}-4\hat{k},\ \vec{b}=-2\hat{i}+4\hat{j}-3\hat{k}\text{ and }\vec{c}=\hat{i}+2\hat{j}-\hat{k},\text{ find }\left\lvert 3\vec{a}-2\vec{b}+4\vec{c}\right\rvert.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: }\overrightarrow{a}=3\widehat{i}-\widehat{j}-4\widehat{k},\;\overrightarrow{b}=-2\widehat{i}+4\widehat{j}-3\widehat{k}\text{ and }\overrightarrow{c}=\widehat{i}+2\widehat{j}-\widehat{k}.$
$\displaystyle \text{Now, }3\overrightarrow{a}-2\overrightarrow{b}+4\overrightarrow{c}=3(3\widehat{i}-\widehat{j}-4\widehat{k})-2(-2\widehat{i}+4\widehat{j}-3\widehat{k})+4(\widehat{i}+2\widehat{j}-\widehat{k})$
$\displaystyle =9\widehat{i}-3\widehat{j}-12\widehat{k}+4\widehat{i}-8\widehat{j}+6\widehat{k}+4\widehat{i}+8\widehat{j}-4\widehat{k}$
$\displaystyle =17\widehat{i}-3\widehat{j}-10\widehat{k}$
$\displaystyle \text{Hence,}$
$\displaystyle |3\overrightarrow{a}-2\overrightarrow{b}+4\overrightarrow{c}|=\sqrt{17^{2}+(-3)^{2}+(-10)^{2}}=\sqrt{289+9+100}=\sqrt{398}$

$\displaystyle \textbf{Question 6: }~\text{If }\overrightarrow{PQ}=3\hat{i}+2\hat{j}-\hat{k} \text{ and the coordinates of }P\text{ are }(1,-1,2), \\\text{ find the coordinates of }Q.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: }\overrightarrow{PQ}=3\widehat{i}+2\widehat{j}-\widehat{k}.$
$\displaystyle \text{Let the position vector of }P(1,-1,2)\text{ be }\overrightarrow{p}\text{ such that }\overrightarrow{p}=\widehat{i}-\widehat{j}+2\widehat{k}$
$\displaystyle \text{and the position vector of }Q(x,y,z)\text{ be }\overrightarrow{q}\text{ such that }\overrightarrow{q}=x\widehat{i}+y\widehat{j}+z\widehat{k}.$
$\displaystyle \text{Therefore,}$
$\displaystyle \overrightarrow{PQ}=\overrightarrow{q}-\overrightarrow{p}$
$\displaystyle \Rightarrow 3\widehat{i}+2\widehat{j}-\widehat{k}=(x\widehat{i}+y\widehat{j}+z\widehat{k})-(\widehat{i}-\widehat{j}+2\widehat{k})$
$\displaystyle \Rightarrow 3\widehat{i}+2\widehat{j}-\widehat{k}=(x-1)\widehat{i}+(y+1)\widehat{j}+(z-2)\widehat{k}$
$\displaystyle \Rightarrow x-1=3,\;y+1=2,\;z-2=-1$
$\displaystyle \Rightarrow x=4,\;y=1,\;z=1$
$\displaystyle \text{Hence, the coordinates of }Q\text{ are }(4,1,1).$

$\displaystyle \textbf{Question 7: }~\text{Prove that the points }\hat{i}-\hat{j},\ 4\hat{i}-3\hat{j}+\hat{k}\text{ and }2\hat{i}-4\hat{j}+5\hat{k}\\ \text{ are the vertices of a right-angled triangle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given the points }\widehat{i}-\widehat{j},\;4\widehat{i}+3\widehat{j}+\widehat{k}\text{ and }2\widehat{i}-4\widehat{j}+5\widehat{k}\text{ are }A,B\text{ and }C\text{ respectively.}$
$\displaystyle \text{Then,}$
$\displaystyle \overrightarrow{AB}=4\widehat{i}+3\widehat{j}+\widehat{k}-(\widehat{i}-\widehat{j})=3\widehat{i}+4\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{BC}=2\widehat{i}-4\widehat{j}+5\widehat{k}-(4\widehat{i}+3\widehat{j}+\widehat{k})=-2\widehat{i}-7\widehat{j}+4\widehat{k}$
$\displaystyle \overrightarrow{CA}=(\widehat{i}-\widehat{j})-(2\widehat{i}-4\widehat{j}+5\widehat{k})=-\widehat{i}+3\widehat{j}-5\widehat{k}$
$\displaystyle \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=3\widehat{i}+4\widehat{j}+\widehat{k}-2\widehat{i}-7\widehat{j}+4\widehat{k}-\widehat{i}+3\widehat{j}-5\widehat{k}=\overrightarrow{0}$
$\displaystyle \text{The given points form the vertices of a triangle.}$
$\displaystyle \text{Now,}$
$\displaystyle |\overrightarrow{AB}|=\sqrt{3^{2}+4^{2}+1^{2}}=\sqrt{9+16+1}=\sqrt{26}$
$\displaystyle |\overrightarrow{BC}|=\sqrt{(-2)^{2}+(-7)^{2}+4^{2}}=\sqrt{4+49+16}=\sqrt{69}$
$\displaystyle |\overrightarrow{CA}|=\sqrt{(-1)^{2}+3^{2}+(-5)^{2}}=\sqrt{1+9+25}=\sqrt{35}$
$\displaystyle |\overrightarrow{AB}|^{2}+|\overrightarrow{CA}|^{2}=26+35=61\neq|\overrightarrow{BC}|^{2}$
$\displaystyle \text{The given triangle is not right-angled.}$

$\displaystyle \textbf{Question 8: }~\text{If the vertices }A,B,C\text{ of a triangle }ABC\text{ are the points with position} \\ \text{vectors} a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k},\ b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k},\ c_{1}\hat{i}+c_{2}\hat{j}+c_{3}\hat{k}\text{ respectively, what are the} \\ \text{vectors determined by its sides? Find the length of these vectors.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given the vertices of a triangle }A,B\text{ and }C\text{ with position vectors }a_{1}\widehat{i}+a_{2}\widehat{j}+a_{3}\widehat{k},\;b_{1}\widehat{i}+b_{2}\widehat{j}+b_{3}\widehat{k}\text{ and }c_{1}\widehat{i}+c_{2}\widehat{j}+c_{3}\widehat{k}\text{ respectively.}$
$\displaystyle \text{Then,}$
$\displaystyle \overrightarrow{AB}=(b_{1}-a_{1})\widehat{i}+(b_{2}-a_{2})\widehat{j}+(b_{3}-a_{3})\widehat{k}$
$\displaystyle \overrightarrow{BC}=(c_{1}-b_{1})\widehat{i}+(c_{2}-b_{2})\widehat{j}+(c_{3}-b_{3})\widehat{k}$
$\displaystyle \overrightarrow{CA}=(a_{1}-c_{1})\widehat{i}+(a_{2}-c_{2})\widehat{j}+(a_{3}-c_{3})\widehat{k}$
$\displaystyle \text{Therefore, the length of these vectors are:}$
$\displaystyle |\overrightarrow{AB}|=\sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}}$
$\displaystyle |\overrightarrow{BC}|=\sqrt{(c_{1}-b_{1})^{2}+(c_{2}-b_{2})^{2}+(c_{3}-b_{3})^{2}}$
$\displaystyle |\overrightarrow{CA}|=\sqrt{(a_{1}-c_{1})^{2}+(a_{2}-c_{2})^{2}+(a_{3}-c_{3})^{2}}$

$\displaystyle \textbf{Question 9: }~\text{Find the vector from the origin }O\text{ to the centroid of the triangle} \\ \text{whose vertices are }(1,-1,2),\ (2,1,3)\text{ and }(-1,2,-1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given the vertices of the triangle }(1,-1,2),\;(2,1,3)\text{ and }(-1,2,-1).$
$\displaystyle \text{Position vectors are}$
$\displaystyle \overrightarrow{a}=\widehat{i}-\widehat{j}+2\widehat{k}$
$\displaystyle \overrightarrow{b}=2\widehat{i}+\widehat{j}+3\widehat{k}$
$\displaystyle \overrightarrow{c}=-\widehat{i}+2\widehat{j}-\widehat{k}$
$\displaystyle \text{The centroid of a triangle is given by }\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}$
$\displaystyle \text{So,}$
$\displaystyle \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}=\frac{\widehat{i}-\widehat{j}+2\widehat{k}+2\widehat{i}+\widehat{j}+3\widehat{k}-\widehat{i}+2\widehat{j}-\widehat{k}}{3}$
$\displaystyle =\frac{2\widehat{i}+2\widehat{j}+4\widehat{k}}{3}=\frac{2}{3}\widehat{i}+\frac{2}{3}\widehat{j}+\frac{4}{3}\widehat{k}$

$\displaystyle \textbf{Question 10: }~\text{Find the position vector of a point }R\text{ which divides the line } \\ \text{segment joining points }P(\hat{i}+2\hat{j}+\hat{k})\text{ and }Q(-\hat{i}+\hat{j}+\hat{k})\text{ in the ratio }2:1:\\ \text{(i) internally}\qquad \text{(ii) externally.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Here }\overrightarrow{a}=\widehat{i}+2\widehat{j}+\widehat{k}\text{ and }\overrightarrow{b}=-\widehat{i}+\widehat{j}+\widehat{k}$
$\displaystyle \text{The position vector of }R\text{ dividing the join of }P\text{ and }Q\text{ internally in the ratio }2:1\text{ is}$
$\displaystyle \overrightarrow{R}=\frac{m\overrightarrow{b}+n\overrightarrow{a}}{m+n}$
$\displaystyle =\frac{2\overrightarrow{b}+1\overrightarrow{a}}{2+1}$
$\displaystyle =\frac{2(-\widehat{i}+\widehat{j}+\widehat{k})+(\widehat{i}+2\widehat{j}+\widehat{k})}{3}$
$\displaystyle =\frac{-\widehat{i}+4\widehat{j}+\widehat{k}}{3}$
$\displaystyle =-\frac{1}{3}\widehat{i}+\frac{4}{3}\widehat{j}+\frac{1}{3}\widehat{k}$
$\displaystyle \text{(ii) }$
$\displaystyle \text{Given: }R\text{ divides the line segment joining the points }P(\widehat{i}+2\widehat{j}+\widehat{k})\text{ and }Q(-\widehat{i}+\widehat{j}+\widehat{k})\text{ in the ratio }2:1\text{ externally.}$
$\displaystyle \text{Therefore, position vector of }R=$
$\displaystyle \frac{2(-\widehat{i}+\widehat{j}+\widehat{k})-1(\widehat{i}+2\widehat{j}+\widehat{k})}{2-1}$
$\displaystyle =-3\widehat{i}+\widehat{k}$

$\displaystyle \textbf{Question 11: }~\text{Find the position vector of the mid-point of the vector } \\ \text{joining the points }P(2\hat{i}-3\hat{j}+4\hat{k})\text{ and }Q(4\hat{i}+\hat{j}-2\hat{k}).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: }P(2\widehat{i}-3\widehat{j}+4\widehat{k})\text{ and }Q(4\widehat{i}+\widehat{j}-2\widehat{k})$
$\displaystyle \text{The position vector of the midpoint of the vector joining these points is}$
$\displaystyle \frac{\text{Position vector of }P+\text{Position vector of }Q}{2}$
$\displaystyle =\frac{(2\widehat{i}-3\widehat{j}+4\widehat{k})+(4\widehat{i}+\widehat{j}-2\widehat{k})}{2}$
$\displaystyle =\frac{6\widehat{i}-2\widehat{j}+2\widehat{k}}{2}$
$\displaystyle =3\widehat{i}-\widehat{j}+\widehat{k}$

$\displaystyle \textbf{Question 12: }~\text{Find the unit vector in the direction of vector }\overrightarrow{PQ},\text{ where }P\text{ and }Q \\ \text{ are the points }(1,2,3)\text{ and }(4,5,6).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ be the position vectors of the points }P(1,2,3)\text{ and }Q(4,5,6).$
$\displaystyle \overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k}$
$\displaystyle \overrightarrow{b}=4\widehat{i}+5\widehat{j}+6\widehat{k}$
$\displaystyle \text{So,}$
$\displaystyle \overrightarrow{PQ}=\overrightarrow{b}-\overrightarrow{a}$
$\displaystyle =4\widehat{i}+5\widehat{j}+6\widehat{k}-\widehat{i}-2\widehat{j}-3\widehat{k}$
$\displaystyle =3\widehat{i}+3\widehat{j}+3\widehat{k}$
$\displaystyle \text{Now,}$
$\displaystyle |\overrightarrow{PQ}|=\sqrt{3^{2}+3^{2}+3^{2}}=\sqrt{9+9+9}=3\sqrt{3}$
$\displaystyle \text{Therefore, unit vector parallel to }\overrightarrow{PQ}\text{ is}$
$\displaystyle \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=\frac{1}{3\sqrt{3}}(3\widehat{i}+3\widehat{j}+3\widehat{k})=\frac{1}{\sqrt{3}}(\widehat{i}+\widehat{j}+\widehat{k})$

$\displaystyle \textbf{Question 13: }~\text{Show that the points }A(2\hat{i}-\hat{j}+\hat{k}),\ B(\hat{i}-3\hat{j}-5\hat{k}),\ \\ C(3\hat{i}-4\hat{j}-4\hat{k})\text{ are the vertices of a right-angled triangle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given the points }A(2\widehat{i}-\widehat{j}+\widehat{k}),\;B(\widehat{i}-3\widehat{j}-5\widehat{k})\text{ and }C(3\widehat{i}-4\widehat{j}-4\widehat{k}).$
$\displaystyle \text{Then, }\overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A$
$\displaystyle =(\widehat{i}-3\widehat{j}-5\widehat{k})-(2\widehat{i}-\widehat{j}+\widehat{k})$
$\displaystyle =-\widehat{i}-2\widehat{j}-6\widehat{k}$
$\displaystyle \overrightarrow{BC}=\text{Position vector of }C-\text{Position vector of }B$
$\displaystyle =(3\widehat{i}-4\widehat{j}-4\widehat{k})-(\widehat{i}-3\widehat{j}-5\widehat{k})$
$\displaystyle =2\widehat{i}-\widehat{j}+\widehat{k}$
$\displaystyle \overrightarrow{CA}=\text{Position vector of }A-\text{Position vector of }C$
$\displaystyle =(2\widehat{i}-\widehat{j}+\widehat{k})-(3\widehat{i}-4\widehat{j}-4\widehat{k})$
$\displaystyle =-\widehat{i}+3\widehat{j}+5\widehat{k}$
$\displaystyle \text{Clearly,}$
$\displaystyle \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{0}$
$\displaystyle \text{Now,}$
$\displaystyle |\overrightarrow{AB}|=\sqrt{(-1)^{2}+(-2)^{2}+(-6)^{2}}=\sqrt{1+4+36}=\sqrt{41}$
$\displaystyle |\overrightarrow{BC}|=\sqrt{2^{2}+(-1)^{2}+1^{2}}=\sqrt{4+1+1}=\sqrt{6}$
$\displaystyle |\overrightarrow{CA}|=\sqrt{(-1)^{2}+3^{2}+5^{2}}=\sqrt{1+9+25}=\sqrt{35}$
$\displaystyle \text{Clearly, }|\overrightarrow{AB}|^{2}=|\overrightarrow{BC}|^{2}+|\overrightarrow{CA}|^{2}$
$\displaystyle \Rightarrow AB^{2}=BC^{2}+CA^{2}$
$\displaystyle \text{So, }A,B,C\text{ form a right angled triangle.}$

$\displaystyle \textbf{Question 14: }~\text{Find the position vector of the mid-point of the vector joining } \\ \text{the points }P(2,3,4)\text{ and }Q(4,1,-2).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{p},\overrightarrow{q}\text{ be the position vectors of the points }P(2,3,4)\text{ and }Q(4,1,-2).$
$\displaystyle \text{Then,}$
$\displaystyle \overrightarrow{p}=2\widehat{i}+3\widehat{j}+4\widehat{k}\text{ and }\overrightarrow{q}=4\widehat{i}+\widehat{j}-2\widehat{k}$
$\displaystyle \text{Therefore, the position vector of the midpoint of the given points is }$
$\displaystyle \frac{\overrightarrow{p}+\overrightarrow{q}}{2}$
$\displaystyle =\frac{(2\widehat{i}+3\widehat{j}+4\widehat{k})+(4\widehat{i}+\widehat{j}-2\widehat{k})}{2}$
$\displaystyle =3\widehat{i}+2\widehat{j}+\widehat{k}$

$\displaystyle \textbf{Question 15: }~\text{Find the value of }x\text{ for which }x(\hat{i}+\hat{j}+\hat{k})\text{ is a unit vector.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }x(\widehat{i}+\widehat{j}+\widehat{k})\text{ is a unit vector.}$
$\displaystyle \therefore \sqrt{x^{2}+x^{2}+x^{2}}=1$
$\displaystyle \Rightarrow \sqrt{3x^{2}}=1$
$\displaystyle \Rightarrow \sqrt{3}|x|=1$
$\displaystyle \Rightarrow |x|=\frac{1}{\sqrt{3}}$
$\displaystyle \Rightarrow x=\pm\frac{1}{\sqrt{3}}$

$\displaystyle \textbf{Question 16: }~\text{If }\vec{a}=\hat{i}+\hat{j}+\hat{k},\ \vec{b}=2\hat{i}-\hat{j}+3\hat{k}\text{ and }\vec{c}=\hat{i}-2\hat{j}+\hat{k},\text{ find a unit vector } \\ \text{parallel to }2\vec{a}-\vec{b}+3\vec{c}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\;\overrightarrow{b}=2\widehat{i}-\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{c}=\widehat{i}-2\widehat{j}+\widehat{k}$
$\displaystyle \therefore 2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}=2(\widehat{i}+\widehat{j}+\widehat{k})-(2\widehat{i}-\widehat{j}+3\widehat{k})+3(\widehat{i}-2\widehat{j}+\widehat{k})$
$\displaystyle =3\widehat{i}-3\widehat{j}+2\widehat{k}$
$\displaystyle \text{A unit vector parallel to }2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}\text{ is given by}$
$\displaystyle \frac{2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}}{|2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}|}$
$\displaystyle =\frac{3\widehat{i}-3\widehat{j}+2\widehat{k}}{\sqrt{3^{2}+(-3)^{2}+2^{2}}}$
$\displaystyle =\frac{3\widehat{i}-3\widehat{j}+2\widehat{k}}{\sqrt{22}}$
$\displaystyle =\frac{3}{\sqrt{22}}\widehat{i}-\frac{3}{\sqrt{22}}\widehat{j}+\frac{2}{\sqrt{22}}\widehat{k}$

$\displaystyle \textbf{Question 17: }~\text{If }\vec{a}=\hat{i}+\hat{j}+\hat{k},\ \vec{b}=4\hat{i}-2\hat{j}+3\hat{k}\text{ and }\vec{c}=\hat{i}-2\hat{j}+\hat{k},\text{ find a vector of magnitude }6\text{ units which is parallel to the vector }2\vec{a}-\vec{b}+3\vec{c}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\;\overrightarrow{b}=4\widehat{i}-2\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{c}=\widehat{i}-2\widehat{j}+\widehat{k}$
$\displaystyle \text{Then,}$
$\displaystyle 2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}=2(\widehat{i}+\widehat{j}+\widehat{k})-(4\widehat{i}-2\widehat{j}+3\widehat{k})+3(\widehat{i}-2\widehat{j}+\widehat{k})$
$\displaystyle =\widehat{i}-2\widehat{j}+2\widehat{k}$
$\displaystyle \text{A unit vector parallel to }2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}\text{ is}$
$\displaystyle \frac{2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}}{|2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}|}$
$\displaystyle =\frac{\widehat{i}-2\widehat{j}+2\widehat{k}}{\sqrt{1^{2}+(-2)^{2}+2^{2}}}$
$\displaystyle =\frac{\widehat{i}-2\widehat{j}+2\widehat{k}}{\sqrt{9}}$
$\displaystyle =\frac{\widehat{i}-2\widehat{j}+2\widehat{k}}{3}$
$\displaystyle \text{Hence, Required vector }=\frac{6}{3}(\widehat{i}-2\widehat{j}+2\widehat{k})=2\widehat{i}-4\widehat{j}+4\widehat{k}$

$\displaystyle \textbf{Question 18: }~\text{Find a vector of magnitude of }5\text{ units parallel to the resultant } \\ \text{of the vectors }\vec{a}=2\hat{i}+3\hat{j}-\hat{k}\text{ and }\vec{b}=\hat{i}-2\hat{j}+\hat{k}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given the position vectors }\overrightarrow{a}=2\widehat{i}+3\widehat{j}-\widehat{k}\text{ and }\overrightarrow{b}=\widehat{i}-2\widehat{j}+\widehat{k}$
$\displaystyle \text{Resultant Vector }=\overrightarrow{a}+\overrightarrow{b}=2\widehat{i}+3\widehat{j}-\widehat{k}+\widehat{i}-2\widehat{j}+\widehat{k}=3\widehat{i}+\widehat{j}$
$\displaystyle \text{So, a unit vector parallel to the resultant vector is }$
$\displaystyle \frac{3\widehat{i}+\widehat{j}}{|3\widehat{i}+\widehat{j}|}=\frac{3\widehat{i}+\widehat{j}}{\sqrt{3^{2}+1^{2}}}=\frac{3\widehat{i}+\widehat{j}}{\sqrt{10}}$
$\displaystyle \text{Hence, required vector }=5\times\frac{3\widehat{i}+\widehat{j}}{\sqrt{10}}=\frac{5}{\sqrt{10}}(3\widehat{i}+\widehat{j})=\sqrt{\frac{5}{2}}(3\widehat{i}+\widehat{j})$

$\displaystyle \textbf{Question 19: }~\text{The two vectors }\hat{j}+\hat{i}\text{ and }3\hat{i}-\hat{j}+4\hat{k}\text{ represent the sides }\overrightarrow{AB}\text{ and }\overrightarrow{AC} \\\text{ respectively of triangle }ABC.\text{ Find the length of the median through }A.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle ABC,\;\overrightarrow{AB}=\widehat{j}+\widehat{k}\text{ and }\overrightarrow{AC}=3\widehat{i}-\widehat{j}+4\widehat{k}$
$\displaystyle \text{Let the position vector of }A\text{ be }(0,0,0).\text{ Then, the position vectors of }B \\ \text{ and }C\text{ are }(0,1,1)\text{ and }(3,-1,4)\text{ respectively.}$
$\displaystyle \text{Suppose }D\text{ be the midpoint of the line segment joining the points }B(0,1,1) \\ \text{ and }C(3,-1,4).$
$\displaystyle \text{Position vector of }D=\frac{(\widehat{j}+\widehat{k})+(3\widehat{i}-\widehat{j}+4\widehat{k})}{2}$
$\displaystyle =\frac{3\widehat{i}+5\widehat{k}}{2}=\frac{3}{2}\widehat{i}+\frac{5}{2}\widehat{k}$
$\displaystyle \text{Now,}$
$\displaystyle |\overrightarrow{AD}|=\left|\left(\frac{3}{2}\widehat{i}+\frac{5}{2}\widehat{k}\right)-(0\widehat{i}+0\widehat{j}+0\widehat{k})\right|$
$\displaystyle =\left|\frac{3}{2}\widehat{i}+\frac{5}{2}\widehat{k}\right|$
$\displaystyle =\sqrt{\left(\frac{3}{2}\right)^{2}+0^{2}+\left(\frac{5}{2}\right)^{2}}$
$\displaystyle =\sqrt{\frac{34}{4}}=\sqrt{\frac{17}{2}}\text{ units}$