Differentiation (Part-38) | S N De

Today we are going to solve Short Answer Type Questions of Exercise -1 of Vector Chapter in S N De Mathematics book. Hopefully, these solutions would come useful for preparations of upcoming examinations. In the previous article , we have already discussed Very Short Answer Type Questions and their solutions. So let's start. To get more solutions of S N Dey mathematics, check out here. [latexpage]

 

$~1.~$ If $~~\vec{a}=\hat{i}+2\hat{j}-\hat{k}~$ and $~\vec{b}=3\hat{i}+\hat{j}-5\hat{k}~$ ,find a unit vector in a direction parallel to vector $~~(\vec{a}-\vec{b}).$

Solution.

$~\vec{a}-\vec{b}\\= (\hat{i}+2\hat{j}-\hat{k})-(3\hat{i}+\hat{j}-5\hat{k})\\= (1-3)\hat{i}+(2-1)\hat{j}+(-1+5)\hat{k} \\= -2\hat{i}+\hat{j}+4\hat{k} \rightarrow(1)\\~\therefore~ |\vec{a}-\vec{b}| \\=\sqrt{(-2)^2+1^2+4^2}\\=\sqrt{4+1+16} \\=\sqrt{21}\rightarrow(2) $

Hence, from $~(1)~$ and $~(2)~,$ we can conclude that a unit vector in a direction parallel to vector $~~(\vec{a}-\vec{b})~$ is :

$~\frac{\vec{a}-\vec{b}}{|\vec{a}-\vec{b}|}=\frac{1}{\sqrt{21}}(-2\hat{i}+\hat{j}+4\hat{k})~\text{(ans.)}$

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 $~2.~$  Let $~\vec{a}=4\hat{i}+3\hat{j}-\hat{k},~\vec{b}=5\hat{i}+2\hat{j}+2\hat{k}, \\~\vec{c}=2\hat{i}-2\hat{j}-3\hat{k}~~\text{and}~~d=4\hat{i}-4\hat{j}+3\hat{k}~;$ 

prove that $~ \vec{b}-\vec{a}~$ and $~ \vec{d}-\vec{c}~$ are parallel and find the ratio of their moduli.

Solution.

$~\vec{b}-\vec{a}\\=(5\hat{i}+2\hat{j}+2\hat{k})-(4\hat{i}+3\hat{j}-\hat{k})\\=(5-4)\hat{i}+(2-3)\hat{j}+(2+1)\hat{k}\\=\hat{i}-\hat{j}+3\hat{k} \rightarrow(1)$

$~\text{Again,}~\vec{d}-\vec{c}\\=(4\hat{i}-4\hat{j}+3\hat{k})-(2\hat{i}-2\hat{j}-3\hat{k})\\=(4-2)\hat{i}+(-4+2)\hat{j}+(3+3)\hat{k}\\=2\hat{i}-2\hat{j}+6\hat{k}\\=2(\hat{i}-\hat{j}+3\hat{k})\\=2(\vec{b}-\vec{a}) ~[\text{By (1)}] \\ ~\therefore~ \vec{d}-\vec{c}=2(\vec{b}-\vec{a})$

Hence, $~(\vec{b}-\vec{a})~~$ and $~~( \vec{d}-\vec{c})~~$ are parallel.

$~\text{Again,}~\vec{d}-\vec{c}=2(\vec{b}-\vec{a}) \\ ~\therefore~ |\vec{d}-\vec{c}|=2|\vec{b}-\vec{a}|\\ \text{or,}~ \frac{|\vec{b}-\vec{a}|}{|\vec{d}-\vec{c}|}=\frac 12 \\ \text{or,}~ |\vec{b}-\vec{a}|:|\vec{d}-\vec{c}|=1:2~\text{(ans.)} $

$3(i)~$ Show that the vectors $~ 2\hat{i}-\hat{j}+\hat{k},~\hat{i}-3\hat{j}-5\hat{k}~$ and $~ -2\hat{i}+3\hat{j}-4\hat{k}~$ are the sides of a right-angled triangle.

Solution.

Let $~\vec{a}=2\hat{i}-\hat{j}+\hat{k},~\vec{b}=\hat{i}-3\hat{j}-5\hat{k},~\vec{c}=-2\hat{i}+3\hat{j}-4\hat{k}.$

$~\therefore~ |\vec{a}|\\=\sqrt{2^2+(-1)^2+1^2}\\=\sqrt{4+1+1}\\=\sqrt{6} \\~\text{or,}~ |\vec{a}|^2=(\sqrt{6})^2=6 \rightarrow(1)$

$\text{Again,}~|\vec{b}|\\=\sqrt{1^2+(-3)^2+(-5)^2}\\=\sqrt{1+9+25}\\=\sqrt{35} \\~\text{or,}~ |\vec{b}|^2=(\sqrt{35})^2=35\rightarrow(2)$

$~\text{and},~ |\vec{c}|\\=\sqrt{(-2)^2+3^2+(-4)^2}\\=\sqrt{4+9+16}\\=\sqrt{29} \\~\text{or,}~ |\vec{c}|^2=(\sqrt{29})^2=29\rightarrow(3)$

Hence, from $~(1),(2),~(3)~$ it follows that

$~~|\vec{b}|^2=|\vec{a}|^2+|\vec{c}|^2.$

So, the given vectors are the sides of a right-angled triangle.

 $~3(ii)~~$ If the position vectors of the points $~A,B,C~$ are $~~3\hat{i}-4\hat{j}-4\hat{k},~~2\hat{i}-\hat{j}+\hat{k}~$ and $~~\hat{i}-3\hat{j}-5\hat{k}~$ respectively. Show that $~ABC~$ is a right angled triangle. $~~[~NCERT~]$

Sol. $~~~~\vec{AB}\\=(\text{position vector of the point}~B~)\\~~~-(\text{position vector of the point}~A~)\\=(2\hat{i}-\hat{j}+\hat{k})-(3\hat{i}-4\hat{j}-4\hat{k})\\=(2-3)\hat{i}+(-1+4)\hat{j}+(1+4)\hat{k}\\=-\hat{i}+3\hat{j}+5\hat{k}$

$~~~~\vec{BC}\\=(\text{position vector of the point}~C~)\\~~~-(\text{position vector of the point}~B~)\\=(\hat{i}-3\hat{j}-5\hat{k})-(2\hat{i}-\hat{j}+\hat{k})\\=(1-2)\hat{i}+(-3+1)\hat{j}+(-5-1)\hat{k}\\=-\hat{i}-2\hat{j}-6\hat{k}$

$~~~~\vec{CA}\\=(\text{position vector of the point}~A~)\\~~~-(\text{position vector of the point}~C~)\\=(3\hat{i}-4\hat{j}-4\hat{k})-(\hat{i}-3\hat{j}-5\hat{k})\\=(3-1)\hat{i}+(-4+3)\hat{j}+(-4+5)\hat{k}\\=2\hat{i}-\hat{j}+\hat{k}$

$~~\therefore~|AB|\\=\sqrt{(-1)^2+3^2+5^2}\\=\sqrt{1+9+25}\\=\sqrt{35},\\~~~~|BC|\\=\sqrt{(-1)^2+(-2)^2+(-6)^2}\\=\sqrt{1+4+36}\\=\sqrt{41},\\~~~~~|CA|\\=\sqrt{2^2+(-1)^2+1^2}\\=\sqrt{4+1+1}\\=\sqrt{6}.$

So, $~|AB|^2\\=(\sqrt{35})^2=35,\\~~|BC|^2\\=(\sqrt{41})^2=41,\\~~|CA|^2=(\sqrt{6})^2=6 \rightarrow(1)$

Hence, from $~(1)~$ we get,

$~~|BC|^2\\=41\\=35+6\\=|AB|^2+|CA|^2 \rightarrow(2)$

So, from $~(2),~$ it is evident that  $~ABC~$ is a right angled triangle.

$~4.~$ If $~\vec{a},\vec{b},~\vec{c}~$ are three given vectors, show that the points having position vectors $~7\vec{a}-\vec{c},~\vec{a}+2\vec{b}+3\vec{c}~$ and $~-2\vec{a}+3\vec{b}+5\vec{c}~$ are collinear.

Sol. Suppose that $~7\vec{a}-\vec{c},~\vec{a}+2\vec{b}+3\vec{c}~$ and $~-2\vec{a}+3\vec{b}+5\vec{c}~$ denote the position vectors of the points $~A,~B,~C~$.

So, $~\vec{AB}\\=(\vec{a}+2\vec{b}+3\vec{c})-(7\vec{a}-\vec{c})\\=-6\vec{a}+2\vec{b}+4\vec{c},\\~\vec{BC}\\=(-2\vec{a}+3\vec{b}+5\vec{c})-(\vec{a}+2\vec{b}+3\vec{c})\\=-3\vec{a}+\vec{b}+2\vec{c}\\~~\text{so that}~~\vec{AB}\\=2(-3\vec{a}+\vec{b}+2\vec{c})\\=2\vec{BC} \rightarrow(1)$

So, from $~(1)~$ we can conclude that $~\vec{AB}~$ and $~\vec{BC}~$ are collinear and so three given points are collinear.

$~5.~$ If $~\vec{a}=\hat{i}+\hat{j}-4\hat{k}~$ and $~~\vec{b}=4\hat{i}-\hat{j}-2\hat{k}~,$ then find

$~(i)~$ a unit vector in the direction of the vector $~(2\vec{a}-\vec{b})~$ and $~(ii)~$ vector and scalar components of the vector $~(2\vec{a}-\vec{b})~$ along the coordinate axes.

Solution $~(i)~$

$~2\vec{a}-\vec{b} \\ =2(\hat{i}+\hat{j}-4\hat{k})-(4\hat{i}-\hat{j}-2\hat{k}) \\= (2-4)\hat{i}+(2+1)\hat{j}+(-8+2)\hat{k} \\ = -2\hat{i}+3\hat{j}-6\hat{k}\rightarrow(1) \\~\therefore~ |2\vec{a}-\vec{b}|\\= \sqrt{(-2)^2+3^2+(-6)^2} \\ =\sqrt{4+9+36} \\= \sqrt{49}\\= 7 \rightarrow(2) $

So, a unit vector in the direction of the vector $~(2\vec{a}-\vec{b})~$ is given by :

$~ \frac{2\vec{a}-\vec{b}}{|2\vec{a}-\vec{b}|}=\frac{-2\hat{i}+3\hat{j}-6\hat{k}}{7} ~ [\text{By (1),(2)}]$

Sol. $~(ii)~~$ The vector component of $~(2\vec{a}-\vec{b})~$ along $~x-$ axis is : $~~-2\hat{i}~,~$ the scalar component of $~(2\vec{a}-\vec{b})~$ along $~x-$ axis is : $~~ -2 .$

The vector component of $~~(2\vec{a}-\vec{b})~~$ along $~y-$ axis is : $~~-2\hat{j} ~,~$ the scalar component of $~~(2\vec{a}-\vec{b})~~$ along $~y-$ axis is : $ ~~3 .$

The vector component of $~(2\vec{a}-\vec{b})~$ along $~~z-$ axis is : $~~-6\hat{k}~~$, the scalar component of $~(2\vec{a}-\vec{b})~$ along $~z-$ axis is : $~~-6 .$

$~6.~$ If the position vectors of the points $~A,B,C~$ are $~ -2\hat{i}+2\hat{j}+2\hat{k}, ~2\hat{i}+3\hat{j}+3\hat{k}~,\\~~-\hat{i}-2\hat{j}+3\hat{k}~~$ respectively, show that $~ABC~$ is an isosceles triangle.

Solution.

$~\vec{AB}\\=(2\hat{i}+3\hat{j}+3\hat{k})-(-2\hat{i}+2\hat{j}+2\hat{k})\\=(2+2)\hat{i}+(3-2)\hat{j}+(3-2)\hat{k}\\=4\hat{i}+\hat{j}+\hat{k},$

$~\vec{BC}\\=(-\hat{i}-2\hat{j}+3\hat{k})-(2\hat{i}+3\hat{j}+3\hat{k})\\=(-1-2)\hat{i}+(-2-3)\hat{j}+(3-3)\hat{k}\\=-3\hat{i}-5\hat{j},$

$~\vec{CA}\\=(-2\hat{i}+2\hat{j}+2\hat{k})-(-\hat{i}-2\hat{j}+3\hat{k})\\=(-2+1)\hat{i}+(2+2)\hat{j}+(2-3)\hat{k}\\=-\hat{i}+4\hat{j}-\hat{k}.$

$~ \therefore~~|\vec{AB}|=\sqrt{4^2+1^2+1^2}=\sqrt{18}, \\~|\vec{BC}|=\sqrt{(-3)^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}, \\~ |\vec{CA}|=\sqrt{(-1)^2+4^2+(-1)^2}=\sqrt{1+16+1}=\sqrt{18}. \\ ~\therefore~ |\vec{AB}|=|\vec{CA}|\rightarrow(1)$

Hence, by $~(1)~$ we can conclude that $~ABC~$ is an isosceles triangle.

$~7(i).~$ If the coordinates of the points $~A,B~$ and $~C~$ are $~(2,6,3)~,(1,2,7)~$ and $~(3,10,-1)~$ respectively, show that by vector method that the points $~A,B~$ and $~C~$ are collinear.

Solution.

According to the problem, the position vectors of the points $~A,B~$ and $~C~$ are $~2\hat{i}+6\hat{j}+3\hat{k}~,~\hat{i}+2\hat{j}+7\hat{k}~$ and $~ 3\hat{i}+10\hat{j}-\hat{k}~$ respectively.

$~~\text{So,}~~\vec{AB}\\=(\hat{i}+2\hat{j}+7\hat{k})-(2\hat{i}+6\hat{j}+3\hat{k})\\=-\hat{i}-4\hat{j}+4\hat{k}\rightarrow(1),$

$~\vec{BC}\\= (3\hat{i}+10\hat{j}-\hat{k})-(\hat{i}+2\hat{j}+7\hat{k})\\=2\hat{i}+8\hat{j}-8\hat{k}\\=-2(-\hat{i}-4\hat{j}+4\hat{k}) \\= -2\vec{AB}~~[\text{By (1)}] \\ ~~\\~~ \therefore~~\vec{BC} = -2\vec{AB} \rightarrow(2)$

Hence, by $~(2)~$ we can conclude that the points $~A,B~$ and $~C~$ are collinear.

$~(ii)~$ The position vectors of three points are

$~(a)~~ -2\hat{i}+3\hat{j}+5\hat{k} ,~~\hat{i}+2\hat{j}+3\hat{k} ,\\~~ ~7\hat{i}-\hat{k}\\~~ \\ ~~(b)~~\hat{i}-2\hat{j}+3\hat{k}, ~~2\hat{i}+3\hat{j}-4\hat{k}, \\~~~-7\hat{j}+10\hat{k}$

In each case show that the three points are collinear.

Solution (a)

let $-2\hat{i}+3\hat{j}+5\hat{k} ,~~\hat{i}+2\hat{j}+3\hat{k},~7\hat{i}-\hat{k}~$ are the position vectors of three points $~A,B,C.$

$ ~\text{So,}~~\vec{AB}\\=(\hat{i}+2\hat{j}+3\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})\\=(1+2)\hat{i}+(2-3)\hat{j}+(3-5)\hat{k}\\=3\hat{i}-\hat{j}-2\hat{k} \rightarrow(1)$

$~~\vec{BC}\\=(7\hat{i}-\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})\\=(7-1)\hat{i}-2\hat{j}+(-1-3)\hat{k}\\=6\hat{i}-2\hat{j}-4\hat{k}\\=2(3\hat{i}-\hat{j}-2\hat{k}) \rightarrow(2)$

So, from $~(1),~(2)~$ we get, $~~\vec{BC}=2\vec{AB}.$

Hence, the three given points are collinear.

Solution (b)

$~\text{Let}~\hat{i}-2\hat{j}+3\hat{k} ,~2\hat{i}+3\hat{j}-4\hat{k},~-7\hat{i}+4\hat{k}~$ are the position vectors of three points $~A,B,C.$

$~\text{So,}~\vec{AB}\\=(2\hat{i}+3\hat{j}-4\hat{k})-(\hat{i}-2\hat{j}+3\hat{k} )\\=(2-1)\hat{i}+(3+2)\hat{j}+(-4-3)\hat{k}\\=\hat{i}+5\hat{j}-7\hat{k} \rightarrow(1)$

$~~\vec{BC}\\=(-7\hat{i}+4\hat{k})-(2\hat{i}+3\hat{j}-4\hat{k})\\=(-7-2)\hat{i}-3\hat{j}+(4+4)\hat{k}\\=-9\hat{i}-3\hat{j}+8\hat{k} \rightarrow(2)$

$~~\vec{AC}\\=(-7\hat{i}+4\hat{k})-(\hat{i}-2\hat{j}+3\hat{k} )\\=(-7-1)\hat{i}+2\hat{j}+(4-3)\hat{k}\\=-8\hat{i}+2\hat{j}+\hat{k} \rightarrow(3)$

So, from $~(1),~(2)~,(3)~$ we get,

$~(\hat{i}+5\hat{j}-7\hat{k})+(-9\hat{i}-3\hat{j}+8\hat{k} ) \\= -8\hat{i}+2\hat{j}+\hat{k} \\ ~\therefore~\vec{AB}+\vec{BC}=\vec{AC}.$

Hence, the three given points are collinear.

$~8.~$ The position vectors of three points are $~\hat{i}+3\hat{j}-2\hat{k},~3\hat{i}-2\hat{j}+\hat{k}~$ and $~-2\hat{i}+\hat{j}+3\hat{k}~;$ show that the points are the vertices of an equilateral triangle.

Solution.

Let $~\hat{i}+3\hat{j}-2\hat{k},~3\hat{i}-2\hat{j}+\hat{k}~$ and $~-2\hat{i}+\hat{j}+3\hat{k}~$ are the position vectors of three points $~A,~B,~C~$ respectively.

$\text{Then,}~\vec{AB}\\=(3\hat{i}-2\hat{j}+\hat{k})-( \hat{i}+3\hat{j}-2\hat{k})\\=2\hat{i}-5\hat{j}+3\hat{k}\rightarrow(1)$

$~\vec{BC}\\=(-2\hat{i}+\hat{j}+3\hat{k})-(3\hat{i}-2\hat{j}+\hat{k})\\=(-2-3)\hat{i}+(1+2)\hat{j}+(3-1)\hat{k}\\=-5\hat{i}+3\hat{j}+2\hat{k}\rightarrow(2)$

$~\vec{CA}\\=(\hat{i}+3\hat{j}-2\hat{k})-(-2\hat{i}+\hat{j}+3\hat{k})\\=(1+2)\hat{i}+(3-1)\hat{j}+(-2-3)\hat{k}\\=3\hat{i}+2\hat{j}-5\hat{k}\rightarrow(3)$

Hence, from $~(1),~(2),~(3)~$ we get,

$~\therefore~|\vec{AB}|\\=\sqrt{2^2+(-5)^2+3^2}\\=\sqrt{4+25+9}\\=\sqrt{38},\\~|\vec{BC}|\\=\sqrt{(-5)^2+3^2+2^2}\\=\sqrt{25+9+4}\\=\sqrt{38},\\~|\vec{CA}|\\=\sqrt{3^2+2^2+(-5)^2}\\=\sqrt{9+4+25}\\=\sqrt{38}\\~\therefore ~|\vec{AB}|=|\vec{BC}|=|\vec{CA}|.$

Hence, the points in question are the vertices of an equilateral triangle.

$~9.~$ If the vectors $~p\hat{i}-5\hat{j}+6\hat{k}~$ and $~2\hat{i}-3\hat{j}-q\hat{k}~$ are collinear , find $~p~$ and $~q.$

Sol. Since the vectors are collinear,

$~p\hat{i}-5\hat{j}+6\hat{k}=\lambda(2\hat{i}-3\hat{j}-q\hat{k}) \\~\text{or,}~p\hat{i}-5\hat{j}+6\hat{k}=(2\lambda)\hat{i}-(3\lambda)\hat{j}+(-q\lambda)\hat{k}\\~\therefore~ p=2\lambda,~5=3\lambda,~6=-q\lambda \rightarrow(1)$

Hence, by $~(1)~$ we get,

$~\lambda=\frac 53 ,~p=2\times \frac 53=\frac{10}{3},\\~q=-\frac{6}{\lambda}=-6 \times \frac 35=-\frac{18}{5}.$

$~10.~$ If the points having position vectors $~\hat{i}+b\hat{j}+c\hat{k},~7\hat{i}+2\hat{j}+6\hat{k}~$ and $~5\hat{i}+2\hat{j}+5\hat{k}~$ are collinear, find the values of $~b~$ and $~c.$

Solution.

Let the position vectors of three points say, $~A,B,C~$ be $~\hat{i}+b\hat{j}+c\hat{k},~7\hat{i}+2\hat{j}+6\hat{k}~$ and $~5\hat{i}+2\hat{j}+5\hat{k}~$

$~\text{So,}~\vec{AB}\\=(7\hat{i}+2\hat{j}+6\hat{k})-(\hat{i}+b\hat{j}+c\hat{k})\\=6\hat{i}+(2-b)\hat{j}+(6-c)\hat{k},$

$~\vec{BC}\\=(5\hat{i}+2\hat{j}+5\hat{k})-(7\hat{i}+2\hat{j}+6\hat{k})\\=-2\hat{i}-\hat{k}.$

Since three points are collinear,

$~\vec{AB}=\lambda(\vec{BC}) \\ \text{or,}~6\hat{i}+(2-b)\hat{j}+(6-c)\hat{k}\\~~=\lambda(-2\hat{i}-\hat{k}) \\ \therefore~ 6\hat{i}+(2-b)\hat{j}+(6-c)\hat{k}\\~~=(-2\lambda)\hat{i}+(-\lambda)\hat{k}\rightarrow(1)$

Now, comparing both sides of $~(1)~$ we get,

$~6=-2\lambda,~2-b=0,~6-c=-\lambda \\ \therefore~ \lambda=-\frac 62=-3,~b=2,\\~ 6-c=-(-3)=3 \Rightarrow c=6-3=3~\text{(ans.)}$

$~11.~$ The position vectors of the points $~A~$ and $~B~$ are $~4\hat{i}-3\hat{j}+5\hat{k}~$ and $~-2\hat{i}+3\hat{j}+2\hat{k}~$ respectively; find

$~(i)~$ the position vector of the middle point of the line-segment $~\overline{AB};$

$~(ii)~$ the position vector of the points of trisection of the line-segment $~\overline{AB}.$

Solution $~(i) :$

The position vector of the middle point of the line-segment $~\overline{AB}~$ is given by :

$~=\frac{(4\hat{i}-3\hat{j}+5\hat{k})+(-2\hat{i}+3\hat{j}+2\hat{k})}{2}\\~~~~=\frac{(4-2)\hat{i}+(-3+3)\hat{j}+(5+2)\hat{k}}{2}\\~~~~=\frac{2\hat{i}+7\hat{k}}{2}\\~~~~=\hat{i}+\frac 72\hat{k}~\text{(ans.)}$

Solution $~(ii).$

Let $~\vec{AB}~$ be trisected by $~C~$ in the ratio $~1:2~$ and trisected by $~D~$ in the ratio $~2:1$

$~\therefore~$ The position vector of $~C~$ is :

$~=\frac{(-2\hat{i}+3\hat{j}+2\hat{k})+2(4\hat{i}-3\hat{j}+5\hat{k})}{1+2}\\~~~~~=\frac{(-2+8)\hat{i}+(3-6)\hat{j}+(2+10)\hat{k}}{3}\\~~~~~=\frac{6\hat{i}-3\hat{j}+12\hat{k}}{3}\\~~~~~=2\hat{i}-\hat{j}+4\hat{k}.$

$~\therefore~$ The position vector of $~D~$ is :

$~=\frac{2(-2\hat{i}+3\hat{j}+2\hat{k})+(4\hat{i}-3\hat{j}+5\hat{k})}{2+1} \\~~~~~=\frac{(-4+4)\hat{i}+(6-3)\hat{j}+(4+5)\hat{k}}{3}\\~~~~~ =\frac{3\hat{j}+9\hat{k}}{3}\\~~~~~=\hat{j}+3\hat{k}~\text{(ans.)}$

$~12.~$ If $~\vec{AB}=2\hat{i}-4\hat{j}+5\hat{k}~$ and $~\vec{BC}=\hat{i}-2\hat{j}-3\hat{k}~$ in parallelogram $~ABCD~$, find a unit vector in direction parallel to the diagonal $~\vec{AC}~$ of the parallelogram.

Solution.

$~\vec{AC}=\vec{AB}+\vec{BC} \\ \Rightarrow ~\vec{AC}=(2\hat{i}-4\hat{j}+5\hat{k})+(\hat{i}-2\hat{j}-3\hat{k}) \\ \Rightarrow~ \vec{AC}=3 \hat{i}-6\hat{j}+2\hat{k} \rightarrow(1) \\~~\\~~ ~\therefore~|\vec{AC}|\\~=\sqrt{3^2+(-6)^2+2^2} \\~= \sqrt{9+36+4}=\sqrt{49}=7\rightarrow(2)$

Therefore, a unit vector in direction parallel to the diagonal $~\vec{AC}~$ of the parallelogram is :

$~=\frac{\vec{AC}}{|\vec{AC}|}=\frac{3 \hat{i}-6\hat{j}+2\hat{k}}{7}~~~[\text{By (1),(2)}]$

$~13.~$ The position vectors of the points $~A~$ and $~B~$ are $~~2\vec{a}+\vec{b}~~$ and $~~\vec{a}-3\vec{b}~~$. If the point $~C~$ divides the line-segment $~\overline{AB}~$ externally in the ratio $~1:2~,$ then find the position vector of the point $~C~$. Show also that $~A~$ is the midpoint of the line-segment $~\overline{CB}.$

Solution.

The position vectors of the points $~A~$ and $~B~$ are $~~2\vec{a}+\vec{b}~~$ and $~~\vec{a}-3\vec{b}~$.

If the point $~C~$ divides the line-segment $~\overline{AB}~$ externally in the ratio $~1:2~,$ then the position vector of the point $~C~$ is given by :

$=\frac{1(\vec{a}-3\vec{b})-2(2\vec{a}+\vec{b})}{1-2}\\~~=\frac{\vec{a}-3\vec{b}-4\vec{a}-2\vec{b}}{-1}\\~~=\frac{-3\vec{a}-5\vec{b}}{-1}\\~~=3\vec{a}+5\vec{b}.$

The position vector of midpoint of the line-segment $~\overline{CB}:$

$=\frac{(3\vec{a}+5\vec{b})+(\vec{a}-3\vec{b})}{2}\\~~~~=\frac{4\vec{a}+2\vec{b}}{2}\\~~~~=2\vec{a}+\vec{b}\\~~~~=\text{the position vector of }~~A.$

Hence, $~~A~$ is the midpoint of the line-segment $~~\overline{CB}.$

$~14.~$ Show that the vector $~\hat{i}+\hat{j}+\hat{k}~$ makes the same angle with positive directions of coordinates axes.

Sol. We know that the unit vector along $~x$-axis is : $~\hat{i}~,$ likewise the unit vectors along $~y$-axis and $~z$-axis are $~\hat{j}~$ and $~\hat{k}~$ respectively.

So, if the vector $~\hat{i}+\hat{j}+\hat{k}~$ makes the angle $~\theta_1~$ with $~x$-axis (i.e. with the vector $~\hat{i})~$, then

$~~\cos \theta_1=\frac{\hat{i} \cdot (\hat{i}+\hat{j}+\hat{k})}{|\hat{i}||\hat{i}+\hat{j}+\hat{k}|}=\frac{1+0+0}{1\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}\rightarrow(1)$

Similarly, if the vector $~\hat{i}+\hat{j}+\hat{k}~$ makes the angle $~\theta_2~$ with $~y$-axis (i.e. with the vector $~\hat{j}~$) ,then

$~~\cos \theta_2=\frac{\hat{j} \cdot (\hat{i}+\hat{j}+\hat{k})}{|\hat{j}||\hat{i}+\hat{j}+\hat{k}|}=\frac{0+1+0}{1\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}\rightarrow(2)$

and the vector $~\hat{i}+\hat{j}+\hat{k}~$ makes the angle $~\theta_3~$ with $~z$-axis (i.e. with the vector $~\hat{k}~$) ,then

$~~\cos \theta_3=\frac{\hat{k} \cdot (\hat{i}+\hat{j}+\hat{k})}{|\hat{k}||\hat{i}+\hat{j}+\hat{k}|}=\frac{0+0+1}{1\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}\rightarrow(3)$

Hence, by $~(1),~(2),~(3)~$ we notice that $~\cos \theta_1=\cos \theta_2=\cos \theta_3~$ and so we can conclude that the vector $~\hat{i}+\hat{j}+\hat{k}~$ makes the same angle with positive directions of coordinates axes.

$~15.~$ Let $~D,~E,~F~$ are the midpoints of the sides $~\overline{BC},~\overline{CA}~$ and $~\overline{AB}~$ of the triangle $~ABC~$ . Prove that $~\vec{AD}+\vec{BE}+\vec{CF}=\vec{0}.$

Sol. Let the position vectors of $~A,~B,~C$ be denoted by $~\vec{a},~\vec{b},~\vec{c}~$ so that the position vectors of $~D,~E,~F~$ are given by $~\frac{\vec{b}+\vec{c}}{2},~\frac{\vec{c}+\vec{a}}{2},~\frac{\vec{a}+\vec{b}}{2}~$ respectively.

$~\therefore~\vec{AD}=\frac{\vec{b}+\vec{c}}{2}-\vec{a},~\vec{BE}=\frac{\vec{c}+\vec{a}}{2}-\vec{b},\\~\vec{CF}=\frac{\vec{a}+\vec{b}}{2}-\vec{c}.$

$~\text{So,}~\vec{AD}+\vec{BE}+\vec{CF}\\=\left(\frac{\vec{b}+\vec{c}}{2}-\vec{a}\right)+\left(\frac{\vec{c}+\vec{a}}{2}-\vec{b}\right)+\left(\frac{\vec{a}+\vec{b}}{2}-\vec{c}\right)\\=\left(\frac{\vec{b}+\vec{c}}{2}+\frac{\vec{c}+\vec{a}}{2}+\frac{\vec{a}+\vec{b}}{2}\right)-(\vec{a}+\vec{b}+\vec{c})\\=\frac{\vec{b}+\vec{c}+\vec{c}+\vec{a}+\vec{a}+\vec{b}}{2}-(\vec{a}+\vec{b}+\vec{c})\\=\frac{2(\vec{a}+\vec{b}+\vec{c})}{2}-(\vec{a}+\vec{b}+\vec{c})\\=(\vec{a}+\vec{b}+\vec{c})-(\vec{a}+\vec{b}+\vec{c})\\=\vec{0}~~(\text{proved})$

$~16.~$ If $~\vec{AO}+\vec{OB}=\vec{BO}+\vec{OC},~$ show that the points $~A,~B~$ and $~C~$ are collinear.

Solution.

$~\vec{AO}+\vec{OB}=\vec{BO}+\vec{OC} \\ \text{or,}~ -\vec{OA}+\vec{OB}=-\vec{OB}+\vec{OC}\\ \text{or,}~ \vec{OB}-\vec{OA}=\vec{OC}-\vec{OB} \\ \therefore ~\text{position vector of }~B \\~~-\text{position vector of }~A\\=\text{position vector of }~C-\\~~\text{position vector of }~B \\ \text{or,}~\vec{AB}=\vec{BC}\rightarrow(1)$

So, by $~(1)~$ we can say that the vectors $~\vec{AB}~$ and $~\vec{BC}~$ have a common point $~C.$

Hence, we can conclude that the points $~A,~B~$ and $~C~$ are collinear.