\displaystyle \textbf{Question 1: }~\text{If }O\text{ is a point in space, }ABC\text{ is a triangle and }D,E,F\text{ are the mid-} \\ \text{points of the sides }BC,CA\text{ and }AB\text{ respectively of the triangle, prove that } \\ \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}.
\displaystyle \text{Answer:}

\displaystyle \text{Let }D,E\text{ and }F\text{ be the midpoints of }BC,CA\text{ and }AB\text{ respectively.}
\displaystyle \text{Therefore,}
\displaystyle \frac{\overrightarrow{OB}+\overrightarrow{OC}}{2}=\overrightarrow{OD}
\displaystyle \overrightarrow{OB}+\overrightarrow{OC}=2\overrightarrow{OD}\ldots(1)
\displaystyle \text{Similarly,}
\displaystyle \overrightarrow{OC}+\overrightarrow{OA}=2\overrightarrow{OE}\ldots(2)
\displaystyle \overrightarrow{OA}+\overrightarrow{OB}=2\overrightarrow{OF}\ldots(3)
\displaystyle \text{Adding }(1),(2)\text{ and }(3).\text{ We get,}
\displaystyle 2(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})=2(\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF})
\displaystyle \Rightarrow \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}
\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 2: }~\text{Show that the sum of three vectors determined by the medians of a } \\ \text{triangle directed from the vertices is zero.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ be the position vectors of the vertices }A,B\text{ and }C\text{ respectively.}
\displaystyle \text{Then we know that the position vector of the centroid }O\text{ of the triangle is } \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}
\displaystyle \text{Therefore, sum of the three vectors }\overrightarrow{OA},\overrightarrow{OB}\text{ and }\overrightarrow{OC}\text{ is}
\displaystyle \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{a}-\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}+\overrightarrow{b}-\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}+\overrightarrow{c}-\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}
\displaystyle =(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})-3\left(\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}\right)
\displaystyle =\overrightarrow{0}
\displaystyle \text{Hence, sum of the three vectors determined by the medians of a triangle directed from} \\ \text{the vertices is zero.}

\displaystyle \textbf{Question 3: }~\text{If }ABCD\text{ is a parallelogram and }P\text{ is the point of intersection of its diagonals. If } \\ O\text{ is the origin of reference, show that }\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=4\overrightarrow{OP}.
\displaystyle \text{Answer:}

\displaystyle \text{Given a parallelogram }ABCD\text{ and }P\text{ is the point of intersection of its diagonals.} \\ \text{We know the diagonals of a parallelogram bisect each other. Therefore,}
\displaystyle \frac{\overrightarrow{OA}+\overrightarrow{OC}}{2}=\overrightarrow{OP}
\displaystyle \overrightarrow{OA}+\overrightarrow{OC}=2\overrightarrow{OP}\ldots(1)
\displaystyle \text{and}
\displaystyle \frac{\overrightarrow{OB}+\overrightarrow{OD}}{2}=\overrightarrow{OP}
\displaystyle \overrightarrow{OB}+\overrightarrow{OD}=2\overrightarrow{OP}\ldots(2)
\displaystyle \text{Adding }(1)\text{ and }(2).\text{ We get,}
\displaystyle \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=4\overrightarrow{OP}

\displaystyle \textbf{Question 4: }~\text{Show that the line segments joining the mid-points of opposite sides of a } \\ \text{quadrilateral bisects each other.}
\displaystyle \text{Answer:}

\displaystyle \text{Join }DB\text{ to form triangle }ABD.
\displaystyle \frac{AS}{SD}=\frac{AP}{PB}
\displaystyle \Rightarrow SP\parallel DB\text{ and }SP=\frac{1}{2}DB
\displaystyle \text{In triangle }BCD
\displaystyle \frac{CR}{RD}=\frac{CQ}{QB}
\displaystyle \Rightarrow RQ\parallel DB\text{ and }RQ=\frac{1}{2}DB
\displaystyle \text{In quadrilateral }PQRS,
\displaystyle SP=RQ\text{ and }SP\parallel RQ
\displaystyle \therefore PQRS\text{ is a parallelogram.}
\displaystyle \text{Diagonals of a parallelogram bisect each other.}
\displaystyle \therefore PR\text{ and }QS\text{ bisect each other.}

\displaystyle \textbf{Question 5: }~\text{If }A,B,C,D\text{ are four points in a plane and }Q\text{ is the point of } \\ \text{intersection of the lines joining the mid-points of }AB\text{ and }CD;\ BC\text{ and }AD. \\ \text{ Show that }\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}+\overrightarrow{PD}=4\overrightarrow{PQ}, \text{ where }P\text{ is any point.}
\displaystyle \text{Answer:}

\displaystyle \text{Let }E,F,G\text{ and }H\text{ be the midpoints of the sides }AB,BC,CD\text{ and } \\ DA\text{ respectively of quadrilateral }ABCD.
\displaystyle \text{By geometry, the figure formed by joining the midpoints }E,F,G\text{ and }H \\ \text{ will be a parallelogram. Hence its diagonals will bisect each other, say at }Q.
\displaystyle \text{Now, }F\text{ is the midpoint of }BC.
\displaystyle \frac{\overrightarrow{PB}+\overrightarrow{PC}}{2}=\overrightarrow{PF}
\displaystyle \Rightarrow \overrightarrow{PB}+\overrightarrow{PC}=2\overrightarrow{PF}\ldots(1)
\displaystyle \text{And, }H\text{ is the midpoint of }AD.
\displaystyle \frac{\overrightarrow{PA}+\overrightarrow{PD}}{2}=\overrightarrow{PH}
\displaystyle \Rightarrow \overrightarrow{PA}+\overrightarrow{PD}=2\overrightarrow{PH}\ldots(2)
\displaystyle \text{Adding }(1)\text{ and }(2).\text{ We get,}
\displaystyle \overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}+\overrightarrow{PD}=2(\overrightarrow{PF}+\overrightarrow{PH})
\displaystyle =2(2\overrightarrow{PQ})=4\overrightarrow{PQ}

\displaystyle \textbf{Question 6: }~\text{Prove by vector method that the internal bisectors of the angles of a } \\ \text{triangle are concurrent.}
\displaystyle \text{Answer:}

\displaystyle \text{Let }ABC\text{ be a triangle and }\overrightarrow{\alpha},\overrightarrow{\beta},\overrightarrow{\gamma}\text{ be the position vectors of the vertices }A,B\text{ and }C\text{ respectively.}
\displaystyle \text{Let }AD,BE\text{ and }CF\text{ be the internal bisectors of }\angle A,\angle B\text{ and }\angle C\text{ respectively.}
\displaystyle \text{We know that }D\text{ divides }BC\text{ in the ratio }AB:AC\text{ that is }c:b.
\displaystyle \text{Then,}
\displaystyle \text{P.V. of }D\text{ is }\frac{c\overrightarrow{\gamma}+b\overrightarrow{\beta}}{c+b}
\displaystyle \text{P.V. of }E\text{ is }\frac{c\overrightarrow{\gamma}+a\overrightarrow{\alpha}}{c+a}
\displaystyle \text{and}
\displaystyle \text{P.V. of }F\text{ is }\frac{a\overrightarrow{\alpha}+b\overrightarrow{\beta}}{a+b}
\displaystyle \text{The point dividing }AD\text{ in the ratio }(b+c):a\text{ is }\frac{a\overrightarrow{\alpha}+b\overrightarrow{\beta}+c\overrightarrow{\gamma}}{a+b+c}
\displaystyle \text{The point dividing }BE\text{ in the ratio }(a+c):b\text{ is }\frac{a\overrightarrow{\alpha}+b\overrightarrow{\beta}+c\overrightarrow{\gamma}}{a+b+c}

\displaystyle \text{The point dividing }CF\text{ in the ratio }(a+b):c\text{ is }\frac{a\overrightarrow{\alpha}+b\overrightarrow{\beta}+c\overrightarrow{\gamma}}{a+b+c}
\displaystyle \text{Since the point }\frac{a\overrightarrow{\alpha}+b\overrightarrow{\beta}+c\overrightarrow{\gamma}}{a+b+c}\text{ lies on all the three internal bisectors }AD,BE\text{ and }CF.
\displaystyle \text{Hence the internal bisectors are concurrent.}

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.