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Class 12 RD Sharma Solutions – Chapter 23 Algebra of Vectors – Exercise 23.3

  • Last Updated : 13 Jan, 2021
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Question 1.  Find the position vector of a point R which divides the line joining the two points P and Q with position vectors \vec{OP}=2\vec{a}+\vec{b}  and  \vec{OQ}=\vec{a}-2\vec{b}   respectively in the ratio 1:2 internally and externally.

Solution:

The point R divides the line joining points P and Q in the ratio 1:2 internally.

The position vector of R =  \frac{\vec{a}-2\vec{b}+2(2\vec{a}+\vec{b})}{1+2}  \frac{5\vec{a}}{3}

Point R divides the line joining P and Q in the ratio 1:2 externally.

The position vector of R = \frac{\vec{a}-2\vec{b}-2(2\vec{a}+\vec{b})}{1-2}



\frac{-3\vec{a}-4\vec{b}}{-1}

3\vec{a}+4\vec{b}

Question 2. Let  \vec{a},\vec{b},\vec{c}   and \vec{d}   be the position vectors of the four distinct points A, B, C, D. If  \vec{b}-\vec{a}=\vec{c}-\vec{d}   then show that ABCD is a parallelogram.                                                                                                         

Solution: 

Given that  are the position vectors of the four distinct points A, B, C, D 

such that \vec{b}-\vec{a} = \vec{c}-\vec{d}

Given that, 

\vec{b}-\vec{a} = \vec{c}-\vec{d}



\vec{AB} = \vec{CD}

So, AB is parallel and equal to DC 

Hence, ABCD is a parallelogram.

Question 3. If \vec{a},\vec{b}   are the position vectors of A, B respectively, find the position vector of a point C in AB produced such that AC = 3AB and that a point D in BA produced such that  BD = 2BA.

Solution: 

Given that \vec{a},\vec{b}  are the position vector of A and B

Let C be a point in AB produced such that AC = 3AB.

From the given data we can say that point C divides the line AB in

Ratio 3:2 externally. So, the position vector of point C can be written as

\vec{c} = \frac{ m\vec{b}-n\vec{a}}{m-n}

\frac{3\vec{b}-2\vec{a}}{3-2}



=  3\vec{b}-2\vec{a}

D be a point in BA produced such that BD = 2BA

It is clear that point D divides the line in 1:2 externally. 

Then the position vector \vec{d}   can be written as

\vec{d} = \frac{m\vec{a}-n{b}}{m-n}

=  \frac{2\vec{a}-\vec{b}}{2-1}

\vec{d} = 2\vec{a}-\vec{b}

Hence  \vec{c} =3\vec{b}-2\vec{a}  and  \vec{d} = 2\vec{a}-\vec{b}

Question 4. Show that the four points A, B, C, D with position vectors  \vec{a},\vec{b},\vec{c}  and \vec{d}  respectively such that 3\vec{a}-2\vec{b}+5\vec{c}-6\vec{d} =\vec{0}  are coplanar. Also, find the position vector of the point of intersection of the lines AC and BD.

Solution:



Given that 3\vec{a}-2\vec{b}+5\vec{c}-6\vec{d} =\vec{0}

3\vec{a}+5\vec{c} = 2\vec{b}+6\vec{d}

Sum of the coefficients on both sides of the given equation is 8

so, divide the equation by 8 on both the sides

\frac{3\vec{a}+5\vec{c}}{8} = \frac{2\vec{b}+6\vec{d}}{8}

\frac{3\vec{a}+5\vec{c}}{3+5} = \frac{2\vec{b}+6\vec{d}}{2+6}

It is clear that the position vector of a point P dividing Ac in the 

Ratio 3:5 is same as that of point P diving BD in the ratio 2:6.

Point P is common to AC and BD. Hence, P is the point of intersection of AC and BD.

Therefore, A, B, C and D are coplanar.



The position vector of point P can be written as 

\frac{3\vec{a}+5\vec{c}}{8}   or \frac{2\vec{b}+6\vec{d}}{8}

Question 5:  Show that the four points P, Q, R, S with position vectors \vec{p},\vec{q},\vec{r}   and \vec{s}  respectively such that 5\vec{p}-2\vec{q}+6\vec{r}-9\vec{s} = 0  are coplanar. Also, find the position vector of the point of intersection of the lines PR and QS.

Solution:

Given that 5\vec{p}-2\vec{q}+6\vec{r}-9\vec{s} = 0

Here  \vec{p}, \vec{q}, \vec{r},  and \vec{s}

are the position vectors of point P, Q, R, S

5\vec{p}+6\vec{r} = 2\vec{q}+9\vec{s}            -(1)

Sum of the coefficients on both the sides of the equation (1) is 11. 

So divide the equation (1) by 11 on both sides.

\frac{5\vec{p}+6\vec{r}}{11} = \frac{2\vec{q}+9\vec{s}}{11}



\frac{5\vec{p}+6\vec{r}}{5+6} = \frac{2\vec{q}+9\vec{s}}{2+9}

It shows that position vector of a point A dividing PR in the ratio of 6:5 and

 QS in the ratio 9:2. So A is the common point to PR and QS.

Therefore, P, Q, R and S are coplanar.

The position vector of point A is given by 

\frac{5\vec{p}+6\vec{q}}{11}  or \frac{2\vec{q}+9\vec{s}}{11}

Question 6: The vertices A, B, C of triangle ABC have respectively position vectors \vec{a},\vec{b},\vec{c}   with respect to a given origin O. Show that the point D where the bisector of  \angle{A}   meets BC has position vector \vec{d}=\frac{\beta\vec{b}+\gamma\vec{c}}{\beta+\gamma}  where \beta=|\vec{c}-\vec{a}| = \gamma=|\vec{b}-\vec{a}|  . Hence deduce that the incentre I has position vector  \frac{\alpha\vec{a}+\beta\vec{b}+\gamma\vec{c}}{\alpha+\beta+\gamma}  where \alpha = |\vec{b}-\vec{c}|

Solution: 

Let ABC be a triangle and the position vectors of A, B, C with respect to some origin say O be 

Let D be the point on BC where the bisector of \angle{A}   meets.

\vec{d} be the position vector of D which divides BC internally in the ratio \beta    



and \gamma   where \beta = |\vec{AC}| and \gamma = {\vec{b}-\vec{a}}    

Thus, \beta=|\vec{c}-\vec{a}| and \gamma=|\vec{b}-\vec{a}|

Therefore, by section formula, the position vector of D is given by

 \vec{OD} = \frac{\beta\vec{b}+\gamma\vec{c}}{\beta+\gamma}

Let \alpha = |\vec{b}-\vec{c}|

Incentre is the concurrent point of angle bisectors.

Thus, Incentre divides the line AD in the ratio \alpha:\beta+\gamma   and 

the position vector of incentre is equal to

\frac{\alpha\vec{a}+\frac{\beta\vec{b}+\gamma\vec{c}}{(\beta+\gamma)}*(\beta+\gamma)}{\alpha+\beta+\gamma} = \frac{\alpha\vec{a}+\beta\vec{b}+\gamma\vec{c}}{\alpha+\beta+\gamma}             

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