\displaystyle \textbf{Question 1: }~\text{Show that the points }A,B,C\text{ with position vectors } \\ \vec{a}-2\vec{b}+3\vec{c},\ 2\vec{a}+3\vec{b}-4\vec{c}\text{ and }-7\vec{b}+10\vec{c}\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }A,B,C\text{ with position vectors }\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c},\;2\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}\text{ and }-7\overrightarrow{b}+10\overrightarrow{c}
\displaystyle \text{Then,}
\displaystyle \overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A
\displaystyle =(2\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c})-(\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c})
\displaystyle =\overrightarrow{a}+5\overrightarrow{b}-7\overrightarrow{c}
\displaystyle \overrightarrow{BC}=\text{Position vector of }C-\text{Position vector of }B
\displaystyle =(-7\overrightarrow{b}+10\overrightarrow{c})-(2\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c})
\displaystyle =-2\overrightarrow{a}-10\overrightarrow{b}+14\overrightarrow{c}
\displaystyle =-2(\overrightarrow{a}+5\overrightarrow{b}-7\overrightarrow{c})
\displaystyle \therefore \overrightarrow{BC}=-2\overrightarrow{AB}
\displaystyle \text{Hence,}
\displaystyle \overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are parallel vectors.}
\displaystyle \text{But }B\text{ is a point common to them.}
\displaystyle \text{So,}
\displaystyle \overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are collinear.}
\displaystyle \text{Hence, points }A,B\text{ and }C\text{ are collinear.}

\displaystyle \textbf{Question 2: }~\text{If }\vec{a},\vec{b},\vec{c}\text{ are non-coplanar vectors, prove that the points } \\ \text{having the following position vectors are collinear:}\\ \text{(i) }\vec{a},\ \vec{b},\ 3\vec{a}-2\vec{b}\qquad \text{(ii) }\vec{a}+\vec{b}+\vec{c},\ 4\vec{a}+3\vec{b},\ 10\vec{a}+7\vec{b}-2\vec{c}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{Given: }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are non-coplanar vectors.}
\displaystyle \text{Let the points be }A,B,C\text{ respectively with position vectors }\overrightarrow{a},\overrightarrow{b},3\overrightarrow{a}-2\overrightarrow{b}.
\displaystyle \text{Then, }\overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A
\displaystyle =\overrightarrow{b}-\overrightarrow{a}
\displaystyle \overrightarrow{BC}=\text{Position vector of }C-\text{Position vector of }B
\displaystyle =(3\overrightarrow{a}-2\overrightarrow{b})-\overrightarrow{b}
\displaystyle =3\overrightarrow{a}-3\overrightarrow{b}
\displaystyle =-3(\overrightarrow{b}-\overrightarrow{a})
\displaystyle \therefore \overrightarrow{BC}=-3\overrightarrow{AB}
\displaystyle \text{So, }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are parallel vectors.}
\displaystyle \text{But }B\text{ is a point common to them.}
\displaystyle \text{Hence, points }A,B\text{ and }C\text{ are collinear.}
\displaystyle \text{(ii) }
\displaystyle \text{Given: }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are non-coplanar vectors.}
\displaystyle \text{Let the points be }A,B,C\text{ respectively with the position vectors }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c},\;4\overrightarrow{a}+3\overrightarrow{b},\;10\overrightarrow{a}+7\overrightarrow{b}-2\overrightarrow{c}.
\displaystyle \text{Then, }\overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A
\displaystyle =(4\overrightarrow{a}+3\overrightarrow{b})-(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})
\displaystyle =3\overrightarrow{a}+2\overrightarrow{b}-\overrightarrow{c}
\displaystyle \overrightarrow{BC}=\text{Position vector of }C-\text{Position vector of }B
\displaystyle =(10\overrightarrow{a}+7\overrightarrow{b}-2\overrightarrow{c})-(4\overrightarrow{a}+3\overrightarrow{b})
\displaystyle =6\overrightarrow{a}+4\overrightarrow{b}-2\overrightarrow{c}
\displaystyle =2(3\overrightarrow{a}+2\overrightarrow{b}-\overrightarrow{c})
\displaystyle \therefore \overrightarrow{BC}=2\overrightarrow{AB}
\displaystyle \text{So, }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are parallel vectors.}
\displaystyle \text{But }B\text{ is a point common to them.}
\displaystyle \text{Hence, }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are collinear.}
\displaystyle \text{Therefore, points }A,B\text{ and }C\text{ are collinear.}

\displaystyle \textbf{Question 3: }~\text{Prove that the points having position vectors }\hat{i}+2\hat{j}+3\hat{k},\ 3\hat{i}+4\hat{j}+7\hat{k},\ -3\hat{i}-2\hat{j}-5\hat{k}\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A,B,C\text{ be the points with position vectors }\widehat{i}+2\widehat{j}+3\widehat{k},\;3\widehat{i}+4\widehat{j}+7\widehat{k},\;-3\widehat{i}-2\widehat{j}-5\widehat{k}.
\displaystyle \text{Then,}
\displaystyle \overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A
\displaystyle =(3\widehat{i}+4\widehat{j}+7\widehat{k})-(\widehat{i}+2\widehat{j}+3\widehat{k})
\displaystyle =2\widehat{i}+2\widehat{j}+4\widehat{k}
\displaystyle \overrightarrow{BC}=\text{Position vector of }C-\text{Position vector of }B
\displaystyle =(-3\widehat{i}-2\widehat{j}-5\widehat{k})-(3\widehat{i}+4\widehat{j}+7\widehat{k})
\displaystyle =-6\widehat{i}-6\widehat{j}-12\widehat{k}
\displaystyle =-3(2\widehat{i}+2\widehat{j}+4\widehat{k})
\displaystyle \therefore \overrightarrow{BC}=-3\overrightarrow{AB}
\displaystyle \text{So, }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are parallel vectors.}
\displaystyle \text{But }B\text{ is a point common to them.}
\displaystyle \text{Hence, }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are collinear.}
\displaystyle \text{Therefore, points }A,B\text{ and }C\text{ are collinear.}

\displaystyle \textbf{Question 4: }~\text{If the points with position vectors }10\hat{i}+3\hat{j},\ 12\hat{i}-5\hat{j}\text{ and }a\hat{i}+11\hat{j}\text{ are collinear, find the value of }a.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A,B,C\text{ be the points with position vectors }10\widehat{i}+3\widehat{j},\;12\widehat{i}-5\widehat{j},\;a\widehat{i}+11\widehat{j}.
\displaystyle \text{Then,}
\displaystyle \overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A
\displaystyle =(12\widehat{i}-5\widehat{j})-(10\widehat{i}+3\widehat{j})
\displaystyle =2\widehat{i}-8\widehat{j}
\displaystyle \overrightarrow{BC}=\text{Position vector of }C-\text{Position vector of }B
\displaystyle =(a\widehat{i}+11\widehat{j})-(12\widehat{i}-5\widehat{j})
\displaystyle =(a-12)\widehat{i}+16\widehat{j}
\displaystyle \text{Since }A,B\text{ and }C\text{ are collinear,}
\displaystyle \overrightarrow{AB}=\lambda\overrightarrow{BC}
\displaystyle \Rightarrow 2\widehat{i}-8\widehat{j}=\lambda\{(a-12)\widehat{i}+16\widehat{j}\}
\displaystyle \Rightarrow 2=\lambda(a-12),\;-8=16\lambda
\displaystyle \Rightarrow \lambda=-\frac{1}{2}
\displaystyle \Rightarrow 2=-\frac{1}{2}(a-12)
\displaystyle \Rightarrow -a+12=4
\displaystyle \Rightarrow a=8

\displaystyle \textbf{Question 5: }~\text{If }\vec{a},\vec{b}\text{ are two non-collinear vectors, prove that the points with } \\ \text{position vectors }\vec{a}+\vec{b},\ \vec{a}-\vec{b}\text{ and }\vec{a}+\lambda\vec{b}\text{ are collinear for all real values of }\lambda.
\displaystyle \text{Answer:}
\displaystyle \text{Given: }\overrightarrow{a},\overrightarrow{b}\text{ are non-collinear vectors.}
\displaystyle \text{Let the position vectors of points }A,B,C\text{ be }\overrightarrow{a}+\overrightarrow{b},\;\overrightarrow{a}-\overrightarrow{b},\;\overrightarrow{a}+\lambda\overrightarrow{b}\text{ respectively.}
\displaystyle \text{Then, }\overrightarrow{AB}=\text{P.V. of }B-\text{P.V. of }A
\displaystyle =(\overrightarrow{a}-\overrightarrow{b})-(\overrightarrow{a}+\overrightarrow{b})
\displaystyle =-2\overrightarrow{b}
\displaystyle \overrightarrow{BC}=\text{P.V. of }C-\text{P.V. of }B
\displaystyle =(\overrightarrow{a}+\lambda\overrightarrow{b})-(\overrightarrow{a}-\overrightarrow{b})
\displaystyle =(\lambda+1)\overrightarrow{b}
\displaystyle \overrightarrow{CA}=\text{P.V. of }A-\text{P.V. of }C
\displaystyle =(\overrightarrow{a}+\overrightarrow{b})-(\overrightarrow{a}+\lambda\overrightarrow{b})
\displaystyle =(1-\lambda)\overrightarrow{b}
\displaystyle \text{Now, the points are collinear if and only if }\overrightarrow{AB}\text{ and }\overrightarrow{CA}\text{ are scalar multiples of }\overrightarrow{BC}
\displaystyle \text{So, }\overrightarrow{AB}=\beta\overrightarrow{BC}
\displaystyle \Rightarrow -2\overrightarrow{b}=\beta(\lambda+1)\overrightarrow{b}
\displaystyle \Rightarrow -2=\beta(\lambda+1)
\displaystyle \Rightarrow \beta=-\frac{2}{\lambda+1}
\displaystyle \therefore \text{for all real values of }\lambda,\text{ the given points are collinear.}

\displaystyle \textbf{Question 6: }~\text{If }\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{BO}+\overrightarrow{OC},\text{ prove that }A,B,C\text{ are collinear points.}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{BO}+\overrightarrow{OC}
\displaystyle \Rightarrow \overrightarrow{AO}-\overrightarrow{BO}=\overrightarrow{OC}-\overrightarrow{OB}
\displaystyle \Rightarrow \overrightarrow{OB}-\overrightarrow{OA}=\overrightarrow{OC}-\overrightarrow{OB}
\displaystyle \Rightarrow \overrightarrow{AB}=\overrightarrow{BC}
\displaystyle \text{Hence, }A,B\text{ and }C\text{ are collinear points.}

\displaystyle \textbf{Question 7: }~\text{Show that the vectors }2\hat{i}-3\hat{j}+4\hat{k}\text{ and }-4\hat{i}+6\hat{j}-8\hat{k}\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Given the position vectors }2\widehat{i}-3\widehat{j}+4\widehat{k}\text{ and }-4\widehat{i}+6\widehat{j}-8\widehat{k}
\displaystyle \text{Let }\overrightarrow{a}=2\widehat{i}-3\widehat{j}+4\widehat{k}\text{ and }\overrightarrow{b}=-4\widehat{i}+6\widehat{j}-8\widehat{k}
\displaystyle \text{Then,}
\displaystyle \overrightarrow{b}=-4\widehat{i}+6\widehat{j}-8\widehat{k}
\displaystyle =-2(2\widehat{i}-3\widehat{j}+4\widehat{k})
\displaystyle =-2\overrightarrow{a}
\displaystyle \text{Hence, }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are collinear.}

\displaystyle \textbf{Question 8: }~\text{If the points }A(m,-1),\ B(2,1)\text{ and }C(4,5)\text{ are collinear, find the value of }m.
\displaystyle \text{Answer:}
\displaystyle \text{The given points are }A(m,-1),\;B(2,1)\text{ and }C(4,5).
\displaystyle \text{Now,}
\displaystyle \overrightarrow{AB}=(2\widehat{i}+\widehat{j})-(m\widehat{i}-\widehat{j})=(2-m)\widehat{i}+2\widehat{j}
\displaystyle \overrightarrow{AC}=(4\widehat{i}+5\widehat{j})-(m\widehat{i}-\widehat{j})=(4-m)\widehat{i}+6\widehat{j}
\displaystyle \text{If }A,B,C\text{ are collinear, then}
\displaystyle \overrightarrow{AB}=\lambda\overrightarrow{AC}
\displaystyle \Rightarrow (2-m)\widehat{i}+2\widehat{j}=\lambda\{(4-m)\widehat{i}+6\widehat{j}\}
\displaystyle \Rightarrow 2-m=\lambda(4-m)\text{ and }2=6\lambda
\displaystyle \Rightarrow \lambda=\frac{1}{3}
\displaystyle \Rightarrow 2-m=\frac{1}{3}(4-m)
\displaystyle \Rightarrow 6-3m=4-m
\displaystyle \Rightarrow 2m=2
\displaystyle \Rightarrow m=1
\displaystyle \text{Thus, the value of }m\text{ is }1.

\displaystyle \textbf{Question 9: }~\text{Show that the points }(3,4),\ (-5,16),\ (5,1)\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the given points be }A(3,4),\;B(-5,16)\text{ and }C(5,1).
\displaystyle \text{Now,}
\displaystyle \overrightarrow{AB}=[(-5\widehat{i}+16\widehat{j})-(3\widehat{i}+4\widehat{j})]=-8\widehat{i}+12\widehat{j}
\displaystyle \overrightarrow{AC}=[(5\widehat{i}+\widehat{j})-(3\widehat{i}+4\widehat{j})]=2\widehat{i}-3\widehat{j}
\displaystyle \text{Clearly,}
\displaystyle \overrightarrow{AB}=-4\overrightarrow{AC}
\displaystyle \text{Therefore,}
\displaystyle \overrightarrow{AB}\text{ and }\overrightarrow{AC}\text{ are parallel vectors. But }A\text{ is a common point of }\overrightarrow{AB}\text{ and }\overrightarrow{AC}
\displaystyle \text{Hence, the given points }(3,4),\;(-5,16)\text{ and }(5,1)\text{ are collinear.}

\displaystyle \textbf{Question 10: }~\text{If the vectors }\vec{a}=2\hat{i}-3\hat{j}\text{ and }\vec{b}=-6\hat{i}+m\hat{j}\text{ are collinear, find the value of }m.
\displaystyle \text{Answer:}
\displaystyle \text{It is given that the vectors }\overrightarrow{a}=2\widehat{i}-3\widehat{j}\text{ and }\overrightarrow{b}=-6\widehat{i}+m\widehat{j}\text{ are collinear.}
\displaystyle \therefore \overrightarrow{b}=\lambda\overrightarrow{a}\text{ for some scalar }\lambda
\displaystyle \Rightarrow -6\widehat{i}+m\widehat{j}=\lambda(2\widehat{i}-3\widehat{j})
\displaystyle \Rightarrow -6\widehat{i}+m\widehat{j}=2\lambda\widehat{i}-3\lambda\widehat{j}
\displaystyle \Rightarrow -6=2\lambda\text{ and }m=-3\lambda
\displaystyle \Rightarrow \lambda=-3
\displaystyle \Rightarrow m=-3(-3)=9
\displaystyle \text{Thus, the value of }m\text{ is }9.

\displaystyle \textbf{Question 11: }~\text{Show that the points }A(1,-2,-8),\ B(5,0,-2)\text{ and }C(11,3,7)\text{ are } \\ \text{collinear, and find the ratio in which }B\text{ divides }AC.
\displaystyle \text{Answer:}
\displaystyle \text{Given points }A(1,-2,-8),\;B(5,0,-2),\;C(11,3,7).
\displaystyle \text{Therefore,}
\displaystyle \overrightarrow{AB}=(5\widehat{i}+0\widehat{j}-2\widehat{k})-(\widehat{i}-2\widehat{j}-8\widehat{k})=4\widehat{i}+2\widehat{j}+6\widehat{k}
\displaystyle \overrightarrow{BC}=(11\widehat{i}+3\widehat{j}+7\widehat{k})-(5\widehat{i}+0\widehat{j}-2\widehat{k})=6\widehat{i}+3\widehat{j}+9\widehat{k}
\displaystyle \overrightarrow{AC}=(11\widehat{i}+3\widehat{j}+7\widehat{k})-(\widehat{i}-2\widehat{j}-8\widehat{k})=10\widehat{i}+5\widehat{j}+15\widehat{k}
\displaystyle \text{Clearly, }\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}
\displaystyle \text{Hence, }A,B,C\text{ are collinear.}
\displaystyle \text{Suppose }B\text{ divides }AC\text{ in the ratio }\lambda:1.\text{ Then the position vector of }B\text{ is}
\displaystyle \left(\frac{11\lambda+1}{\lambda+1}\right)\widehat{i}+\left(\frac{3\lambda-2}{\lambda+1}\right)\widehat{j}+\left(\frac{7\lambda-8}{\lambda+1}\right)\widehat{k}
\displaystyle \text{But the position vector of }B\text{ is }5\widehat{i}+0\widehat{j}-2\widehat{k}
\displaystyle \Rightarrow \frac{11\lambda+1}{\lambda+1}=5,\;\frac{3\lambda-2}{\lambda+1}=0,\;\frac{7\lambda-8}{\lambda+1}=-2
\displaystyle \Rightarrow 11\lambda+1=5\lambda+5,\;3\lambda-2=0,\;7\lambda-8=-2\lambda-2
\displaystyle \Rightarrow 6\lambda=4,\;3\lambda=2,\;9\lambda=6
\displaystyle \Rightarrow \lambda=\frac{2}{3}

\displaystyle \textbf{Question 12: }~\text{Using vectors show that the points }A(-2,3,5),\ B(7,0,-1),\ C(-3,-2,-5) \\ \text{ and }D(3,4,7)\text{ are such that }AB\text{ and }CD\text{ intersect at the point }P(1,2,3).
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \overrightarrow{AP}=\text{Position vector of }P-\text{Position vector of }A
\displaystyle =(\widehat{i}+2\widehat{j}+3\widehat{k})-(-2\widehat{i}+3\widehat{j}+5\widehat{k})
\displaystyle =3\widehat{i}-\widehat{j}-2\widehat{k}
\displaystyle \overrightarrow{PB}=\text{Position vector of }B-\text{Position vector of }P
\displaystyle =(7\widehat{i}+0\widehat{j}-\widehat{k})-(\widehat{i}+2\widehat{j}+3\widehat{k})
\displaystyle =6\widehat{i}-2\widehat{j}-4\widehat{k}
\displaystyle \text{Since }\overrightarrow{PB}=2\overrightarrow{AP},\text{ vectors }\overrightarrow{PB}\text{ and }\overrightarrow{AP}\text{ are collinear. But }P\text{ is a common point to them.}
\displaystyle \text{Hence, }P,A,B\text{ are collinear points.}
\displaystyle \text{Now,}
\displaystyle \overrightarrow{CP}=(-3\widehat{i}-2\widehat{j}-5\widehat{k})-(\widehat{i}+2\widehat{j}+3\widehat{k})
\displaystyle =-4\widehat{i}-4\widehat{j}-8\widehat{k}
\displaystyle \overrightarrow{PD}=(\widehat{i}+2\widehat{j}+3\widehat{k})-(3\widehat{i}+4\widehat{j}+7\widehat{k})
\displaystyle =-2\widehat{i}-2\widehat{j}-4\widehat{k}
\displaystyle \text{Thus, }\overrightarrow{CP}=2\overrightarrow{PD}
\displaystyle \text{So the vectors }\overrightarrow{CP}\text{ and }\overrightarrow{PD}\text{ are collinear. But }P\text{ is a common point to them.}
\displaystyle \text{Hence, }C,P,D\text{ are collinear points.}
\displaystyle \text{Thus }A,B,C,D\text{ and }P\text{ are points such that }A,P,B\text{ and }C,P,D\text{ are two sets of collinear points.}
\displaystyle \text{Hence, }AB\text{ and }CD\text{ intersect at point }P.

\displaystyle \textbf{Question 13: }~\text{Using vectors, find the value of }\lambda\text{ such that the points }(\lambda,-10,3),\  \\ (1,-1,3)\text{ and }(3,5,3)\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the given points be }A(\lambda,-10,3),\;B(1,-1,3)\text{ and }C(3,5,3).
\displaystyle \overrightarrow{AB}=(\widehat{i}-\widehat{j}+3\widehat{k})-(\lambda\widehat{i}-10\widehat{j}+3\widehat{k})=(1-\lambda)\widehat{i}+9\widehat{j}
\displaystyle \overrightarrow{AC}=(3\widehat{i}+5\widehat{j}+3\widehat{k})-(\lambda\widehat{i}-10\widehat{j}+3\widehat{k})=(3-\lambda)\widehat{i}+15\widehat{j}
\displaystyle \text{If the points }A,B,C\text{ are collinear, then}
\displaystyle \overrightarrow{AB}=k\overrightarrow{AC}\text{ for some scalar }k
\displaystyle \Rightarrow (1-\lambda)\widehat{i}+9\widehat{j}=k\{(3-\lambda)\widehat{i}+15\widehat{j}\}
\displaystyle \Rightarrow 1-\lambda=k(3-\lambda)\text{ and }9=15k
\displaystyle \Rightarrow k=\frac{3}{5}
\displaystyle \Rightarrow 1-\lambda=\frac{3}{5}(3-\lambda)
\displaystyle \Rightarrow 5-5\lambda=9-3\lambda
\displaystyle \Rightarrow 2\lambda=-4
\displaystyle \Rightarrow \lambda=-2
\displaystyle \text{Thus, the value of }\lambda\text{ is }-2.

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