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Class 12 RD Sharma Solutions – Chapter 23 Algebra of Vectors – Exercise 23.4

  • Last Updated : 21 Feb, 2021
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Question 1. If O is a point in space, ABC is a triangle, and D, E, F are the mid-points of the sides BC, CA, and AB respectively of the triangle, prove that

\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}

Solution:

In the △ABC, D, E, F are the mid points of the sides of BC, CA and AB and O is any point in space.

Let us considered \vec{a},\ \ \vec{b},\ \ \vec{c},\ \ \vec{d},\ \ \vec{e},\ \ \vec{f}  be the position vector of point A, B, C, D, E, F with respect to O.

Therefore, \overrightarrow{OA}=\vec{a},\ \ \overrightarrow{OB}=\vec{b},\ \ \overrightarrow{OC}=\vec{c}\\ \overrightarrow{OD}=\vec{d},\ \ \overrightarrow{OE}=\vec{e},\ \ \overrightarrow{OF}=\vec{f}



So, according to the mid-point formula

 \vec{d}=\frac{\vec{b}+\vec{c}}{2}\\ \vec{e}=\frac{\vec{a}+\vec{c}}{2}\\ \vec{f}=\frac{\vec{a}+\vec{b}}{2}

\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}=\vec{d}+\vec{e}+\vec{f}\\ =\frac{\vec{b}+\vec{c}}{2}+\frac{\vec{a}+\vec{c}}{2}+\frac{\vec{a}+\vec{b}}{2}\\ =\frac{\vec{b}+\vec{c}+\vec{a}+\vec{c}+\vec{a}+\vec{b}}{2}\\ =\frac{2(\vec{a}+\vec{b}+\vec{c})}{2}\\ =\vec{a}+\vec{b}+\vec{c}\\ =\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}

Hence, proved

\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}

Question 2. Show that the sum of three vectors determined by the medians of a triangle directed from the vertices is zero.

Solution:

Let us considered ABC is triangle, so the position vector of A, B and C are \vec{a},\ \ \vec{b}\ and\ \ \vec{c}   



Hence, AD, BE, CF are medium, so, D, E and F are mid points of line BC, AC, and AB.

Now, using mid point formula we get

Position vector of D = \frac{\vec{b}+\vec{c}}{2}                                            

Position vector of E = \frac{\vec{c}+\vec{a}}{2}                                               

Position vector of F = \frac{\vec{a}+\vec{b}}{2}     

Now, add all the three median                                          

\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{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)\\ =\frac{\vec{b}+\vec{c}-2\vec{a}}{2}+=\frac{\vec{c}+\vec{a}-2\vec{b}}{2}+=\frac{\vec{a}+\vec{b}-2\vec{c}}{2}\\ =\frac{\vec{b}+\vec{c}-2\vec{a}+\vec{c}+\vec{a}-2\vec{b}+\vec{a}+\vec{b}-2\vec{c}}{2}\\ =\frac{2\vec{b}+2\vec{c}+2\vec{a}-2\vec{b}-2\vec{a}-2\vec{c}}{2}\\ =\frac{\vec{0}}{2}\\ =\vec{0}\\ \therefore\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\vec{0}

Hence, proved that the sum of the three vectors determined by the medians 

of a triangle directed from the vertices is zero.



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

Solution:

Given that ABCD is a parallelogram, P is the point of intersection of diagonals and O be the point of reference.

So, by using triangle law in △AOP, we get

\overrightarrow{OP}+\overrightarrow{PA}=\overrightarrow{OA}          -(1)

By using triangle law in △OBP, we get

\overrightarrow{OP}+\overrightarrow{PB}=\overrightarrow{OB}         -(2)

By using triangle law in △OPC, we get

\overrightarrow{OP}+\overrightarrow{PC}=\overrightarrow{OC}         -(3)

By using triangle law in △OPD, we get



\overrightarrow{OP}+\overrightarrow{PD}=\overrightarrow{OD}         -(4)

Now, on adding equation (1), (2), (3) and (4), we get

\overrightarrow{OP}+\overrightarrow{PA}+\overrightarrow{OP}+\overrightarrow{PB}+\overrightarrow{OP}+\overrightarrow{PC}+\overrightarrow{OP}+\overrightarrow{PD}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\\ 4\overrightarrow{OP}+\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}+\overrightarrow{PD}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\\ 4\overrightarrow{OP}+\overrightarrow{PA}+\overrightarrow{PB}-\overrightarrow{PA}-\overrightarrow{PB}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\\  Since, \overrightarrow{PC}=-\overrightarrow{PA}\ and\ \overrightarrow{PD}=-\overrightarrow{PB}\ as\ P\ is\ mid\ point\ of\ AC,\ BD\\ 4\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}

Question 4. Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other.

Solution:

Let us considered ABCD be a quadrilateral and P, Q, R, S be the mid points of sides AB, BC, CD and DA.

So, the position vector of A, B, C and D be \vec{a},\ \vec{b},\ \vec{c}\ and\ \vec{d}.

Using the mid point formula 

Position vector of P = \frac{\vec{a}+\vec{b}}{2}

Position vector of Q = \frac{\vec{b}+\vec{c}}{2}



Position vector of R = \frac{\vec{c}+\vec{d}}{2}

Position vector of S = \frac{\vec{d}+\vec{a}}{2}

Position vector of \overrightarrow{PQ} = Position vector of Q – Position vector of P

=\left(\frac{\vec{b}+\vec{c}}{2}\right)-\left(\frac{\vec{a}+\vec{b}}{2}\right)\\ =\frac{\vec{b}+\vec{c}-\vec{a}-\vec{b}}{2}\\ =\frac{\vec{c}+\vec{a}}{2}          -(1)

Position vector of \overrightarrow{SR} = Position vector of R – Position vector of S

=\left(\frac{\vec{c}+\vec{d}}{2}\right)-\left(\frac{\vec{a}+\vec{d}}{2}\right)\\ =\frac{\vec{c}+\vec{d}-\vec{a}-\vec{d}}{2}\\ =\frac{\vec{c}-\vec{a}}{2}        -(2)

From eq(1) and (2),

\overrightarrow{PQ}=\overrightarrow{SR}

So, PQRS is a parallelogram and PR bisects QS             -(as diagonals of parallelogram)

Hence, proved that the line segments joining the mid-points of 



opposite sides of a quadrilateral bisect each other. 

Question 5. ABCD are four points in a plane and Q is the point of intersection of the lines joining the mid-points of AB and CD; BC and AD. Show that \overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}+\overrightarrow{PD}=4\overrightarrow{PQ}  where P is any point.

Solution:

Let us considered the position vector of the points A, B, C and D are\vec{a},\ \ \vec{b},\ \ \vec{c},\ \ \vec{d}

Using the mid-point formula, we get

Position vector of AB = \frac{\vec{a}+\vec{b}}{2}

Position vector of BC = \frac{\vec{b}+\vec{c}}{2}

Position vector of CD = \frac{\vec{c}+\vec{d}}{2}

Position vector of DA = \frac{\vec{a}+\vec{d}}{2}

It is given that Q is the mid point of the line joining the mid points of AB and CD, so



Q=\frac{\frac{\vec{a}+\vec{b}}{2}+\frac{\vec{c}+\vec{d}}{2}}{2}\\ =\frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}

Now, let us assume \vec{p} be the position vector of P.

So,

\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}+\overrightarrow{PD}\\ =\vec{a}-\vec{p}+\vec{b}-\vec{p}+\vec{c}-\vec{p}+\vec{d}-\vec{p}\\ =(\vec{a}+\vec{b}+\vec{c}+\vec{d})-4\vec{p}\\ =4\left(\frac{\vec{a}+\vec{b}+\vec{c}+\vec{d} }{4}-p\right)\\ =4\overrightarrow{PQ}

Hence proved

Question 6. Prove by vector method that the internal bisectors of the angles of a triangle are concurrent.

Solution:

In triangle ABC, let us assume that the position vectors of the vertices of the triangle are A(\vec{a}),  B(\vec{b})\ and\ C(\vec{c}).

Length of the sides:

BC = x



AC = y

AB = z

In triangle ABC, the internal bisector divides the opposite side in the ratio of the sides containing the angles.

Since AD is the internal bisector of the ∠ABC, so

BD/DC = AB/AC = z/y          -(1)

Therefore, position vector of D = \frac{z\vec{c}+y\vec{b}}{y+z}   

Let the internal bisector intersect at point I.

ID/AI = BD/AB         -(2)

BD/DC = z/y

Therefore,

CD/BD = z/y

(CD + BD)/BD = (y + z)/z

BC/BD = (y + z)/z

BD = ln/y + z         -(3)

So, from eq(2) and (3), we get

ID/AI = ln/(y +z)

Therefore,

Position vector of I = \frac{\left(\frac{zc+yb}{y+z}\right)(y+z)+la}{l+y+z}=\frac{la+yb+zc}{l+y+z}

Similarly, we can also prove that I lie on the internal bisectors of ∠B and ∠C. 

Hence, the bisectors are concurrent.

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