Skip to content
Related Articles

Related Articles

Improve Article

Class 12 RD Sharma Solutions – Chapter 23 Algebra of Vectors – Exercise 23.2

  • Last Updated : 08 May, 2021

Question 1. If P, Q, and R are three collinear points such that \overrightarrow{PQ}=\overrightarrow{a} and \overrightarrow{QR}=\overrightarrow{b}.  Find the vector \overrightarrow{PR}

Solution:

According to the question, given that 

Points P, Q, and R are collinear. 

Also, \overrightarrow{PQ}=\overrightarrow{a}and \overrightarrow{QR}=\overrightarrow{b}

So, 



\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR}\\ =\overrightarrow{a}+\overrightarrow{b}

\overrightarrow{PR}=\overrightarrow{a}+\overrightarrow{b}

Question 2. Given condition that three vectors \overrightarrow{a},\overrightarrow{b},  and \space \overrightarrow{c}  form the three sides of a triangle. What are other possibilities?

Solution:

According to the question, given that \overrightarrow{a},\overrightarrow{b}and\space \overrightarrow{c}     are three sides of a triangle ABC.

\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\\ =\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}

=\overrightarrow{AC}+\overrightarrow{CA}             [since \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}]

=\overrightarrow{AC}-\overrightarrow{AC}                  [since \overrightarrow{CA}=\overrightarrow{AC}]

So, \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}



As we know that if vectors are represented in magnitude and direction by the two sides 

of triangle taken is same order, then their sum is represented by the third side taken in reverse order.

So, 

\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}

or 

\overrightarrow{a+}\overrightarrow{c}=\overrightarrow{b}\\ \overrightarrow{b}+\overrightarrow{c}=\overrightarrow{a}

Question 3. If \overrightarrow{a} and \overrightarrow{b} are two non- collinear vectors having the same initial point. What are the vectors represented by \overrightarrow{a}+\overrightarrow{b} and \overrightarrow{a}-\overrightarrow{b}?

Solution:

According to the question, given that \overrightarrow{a}     and \overrightarrow{b}

are two non-collinear vectors having the same initial point.

So, let us considered \overrightarrow{a}=\overrightarrow{AB}\space and \overrightarrow{b}=\overrightarrow{AD}



Now we draw a parallelogram named as ABCD 

Using the properties of parallelogram, we get

\overrightarrow{BC}=\overrightarrow{b}\space and \space \overrightarrow{DC}=\overrightarrow{a}

In ∆ABC,

Using the triangle law, we get

\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}

\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{AC}     …….(i)

In ∆ABD,

Using the triangle law, we get

\overrightarrow{AD}+\overrightarrow{DB}=\overrightarrow{AB}\\ \overrightarrow{b}+\overrightarrow{DB}=\overrightarrow{a}



\overrightarrow{DB}=\overrightarrow{a}-\overrightarrow{b}     …….(ii)

On solving equation (i) and (ii), we get 

\overrightarrow{a}+\overrightarrow{b} and \overrightarrow{a}-\overrightarrow{b}       

are diagonals of a parallelogram whose adjacent sides are \overrightarrow{a} and \overrightarrow{b}

Question 4. If \overrightarrow{a} is a vector and m is a scalar such that m\vec{a}=0, then what are the alternatives for m and \overrightarrow{a}?

Solution:

According to the question, given that m is a scalar and \overrightarrow{a} is a vector such that

\overrightarrow{ma}=\overrightarrow{o}

m(a_1\overrightarrow{i}+b_1\overrightarrow{j}+c_1\overrightarrow{k}=)                              [since let \overrightarrow{a}=a_1\overrightarrow{i}+b_1\overrightarrow{j}+c_1\overrightarrow{k}      ]

ma_1\overrightarrow{i}+mb_1\overrightarrow{j}+mc_1\overrightarrow{k}=0*\overrightarrow{i}*+0*\overrightarrow{j}+0*\overrightarrow{k}

Now on comparing the coefficients of \overrightarrow{i},\overrightarrow{j},\overrightarrow{k} of LHS and RHS, we get



ma1 = 0 ⇒ m = 0 or a1 = 0      …….(i)

mb1 = 0 ⇒ m = 0 or b1 = 0          …….(ii)

mc1 = 0 ⇒ m = 0 or c1 = 0         …….(iii)

Now from eq (i), (ii) and (iii), we get

m = 0 or a1 = b1 = c1 = 0

m = 0 or \overrightarrow{a}=a_1\overrightarrow{i}+b_1\overrightarrow{j}+c_1\overrightarrow{j}=0

m = 0 or \overrightarrow{a}=0

Question 5. If \overrightarrow{a},\overrightarrow{b}  are two vectors, then write the truth value of the following statement:

(i) \overrightarrow{a}=-\overrightarrow{b}   \begin{vmatrix} \overrightarrow{a} \end{vmatrix}=\begin{vmatrix} \overrightarrow{b} \end{vmatrix}

(ii)\begin{vmatrix} \overrightarrow{a} \end{vmatrix}=\begin{vmatrix} \overrightarrow{b} \end{vmatrix}⇒ \overrightarrow{a} =±\overrightarrow{b}     

(iii)\begin{vmatrix} \overrightarrow{a} \end{vmatrix}=\begin{vmatrix} \overrightarrow{b} \end{vmatrix} ⇒\overrightarrow{a} =\overrightarrow{b}

Solution:

(i) Let us assume \overrightarrow{a}=a_1\overrightarrow{i}+b_1\overrightarrow{j}+c_1\overrightarrow{k}\\ \overrightarrow{b}=a_2\overrightarrow{i}+b_2\overrightarrow{j}+c_2\overrightarrow{k}

Given that, a = -b



So, 

a_1\overrightarrow{i}+b_1\overrightarrow{j}+c_1\overrightarrow{k}=-a_2\overrightarrow{i}-b_2\overrightarrow{j}-c_2\overrightarrow{k}

Now on comparing the coefficients of i, j, k in LHS and RHS, we get

a1 = a2         …….(i)

b1 = b2         …….(ii)

c1 = c2         …….(iii)

\begin{vmatrix} \overrightarrow{a} \end{vmatrix}=\sqrt{a_1^2+b_1^2+c_1^2}

From eq(i), (ii), and (iii),

\begin{vmatrix} \overrightarrow{a} \end{vmatrix}=\sqrt{(-a_2)^2+(-b_2)^2+(-c_2)^2}\\ \begin{vmatrix} \overrightarrow{a} \end{vmatrix}=\sqrt{a_2^2+b_2^2+c_2^2}\\ \begin{vmatrix} \overrightarrow{a} \end{vmatrix}=\begin{vmatrix} \overrightarrow{b} \end{vmatrix}

(ii) Given a and b are two vectors such that \begin{vmatrix} \overrightarrow{a} \end{vmatrix}=\begin{vmatrix} \overrightarrow{b} \end{vmatrix}



So, it means the magnitude of vector \overrightarrow{a}   is equal to the magnitude 

of vector \overrightarrow{b}   , but we cannot find the direction of the vector.

Hence, it is false that

\begin{vmatrix} \overrightarrow{a} \end{vmatrix}=\begin{vmatrix} \overrightarrow{b} \end{vmatrix}⇒\overrightarrow{a} =±\overrightarrow{b}

(iii) Given for any vector \overrightarrow{a} and  \overrightarrow{b}

are equal but we cannot find the direction of the vector of \overrightarrow{a} \space and   \space \overrightarrow{b}     

So, it is false.

Question 6. ABCD is a quadrilateral. Find the sum of the vectors \overrightarrow{BA},\overrightarrow{BC},\overrightarrow{CD},  and \overrightarrow{DA}  .

Solution:

According to the question,

ABCD is a quadrilateral.



so, 

In ∆ADC,  

By using triangle law, we get

\overrightarrow{CD}+\overrightarrow{DA}=\overrightarrow{CA}   ……(i)

In ∆ABC, 

By using triangle law, we get

\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{BA}    ……(ii)

Now put the value of \overrightarrow{CA}   in equation (ii), we get

\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}=\overrightarrow{BA}

Now on adding \overrightarrow{BA}   on both sides,

\overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{ED}+\overrightarrow{AC}=3\overrightarrow{AC}    

\overrightarrow{BA}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}=\overrightarrow{BA}+\overrightarrow{BA}\\ \overrightarrow{BA}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}=2\overrightarrow{BA}

Question 7. ABCDE is a pentagon, prove that

(i)\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EA}=0

(ii)\overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{BC}+\overrightarrow{ED}+\overrightarrow{AC}=3\overrightarrow{AC}

Solution:

(i) According to the question, 

ABCDE is a pentagon,

So, 

\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EA}=0 =(\overrightarrow{AB}+\overrightarrow{BC})+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EA}=0

Using the law of triangle \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC} , we get

\overrightarrow{AC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EA}                  

=(\overrightarrow{AC}+\overrightarrow{CD})+\overrightarrow{DE}+\overrightarrow{EA}



=(\overrightarrow{AD})+\overrightarrow{DE}+\overrightarrow{EA}

Using triangle law ,\overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}] , we get

=\overrightarrow{AD}+\overrightarrow{DA}

=\overrightarrow{AD}-(-\overrightarrow{AD})

= 0

\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EA}=0

Hence Proved

(ii) According to the question, 

ABCDE is a pentagon,

So, 

\overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{ED}+\overrightarrow{AC}\\ =(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{AE}+\overrightarrow{DC}+\overrightarrow{ED}+\overrightarrow{AC})

=\overrightarrow{AC}+\overrightarrow{DC}+(\overrightarrow{AE}+\overrightarrow{ED})+\overrightarrow{AC}    

Using triangle law,\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}   , we get

=\overrightarrow{AC}+\overrightarrow{DC}+(\overrightarrow{AD})+\overrightarrow{AC}\\ =\overrightarrow{AC}+\overrightarrow{DC}-\overrightarrow{DA}+\overrightarrow{AC}\\ =\overrightarrow{AC}+\overrightarrow{DC}+\overrightarrow{AD}+\overrightarrow{AC}\\ =\overrightarrow{AC}+\overrightarrow{AC}+\overrightarrow{AC}\\ =3\overrightarrow{AC}

Hence Proved

Question 8. Prove that the sum of all vectors drawn from the centre of a regular octagon to its vertices is the zero vector.

Solution:

Let us assume O be the centre of a regular octagon, as we know that the 

centre of a regular octagon bisects all the diagonals passing through it.

So, 

\overrightarrow{OA}=\overrightarrow{OE}       …….(i)



\overrightarrow{OB}=-\overrightarrow{OF}     …….(ii)

\overrightarrow{OC}=-\overrightarrow{OG}    …….(iii)

\overrightarrow{OD}=-\overrightarrow{OH}   [Tex]\overrightarrow{OD}=-\overrightarrow{OH}   [/Tex]     …….(iv)

Now on adding equation (i), (ii), and (iv), we get

\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=-\overrightarrow{OE}-\overrightarrow{OF}-\overrightarrow{OG}-\overrightarrow{OH}\\ \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=-\overrightarrow{OE}-\overrightarrow{OF}-\overrightarrow{OG}-\overrightarrow{OH}\\ \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=-(\overrightarrow{OE}+\overrightarrow{OF}+\overrightarrow{OG}+\overrightarrow{OH})\\ \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}+\overrightarrow{OG}+\overrightarrow{OH}=0

Hence proved

Question 9. If P is a point and ABCD is quadrilateral \overrightarrow{AP}+\overrightarrow{PB}+\overrightarrow{PD}=\overrightarrow{PC} and, show that ABCD is a parallelogram.

Solution:

According to the question

 \overrightarrow{AP}+\overrightarrow{PB}+\overrightarrow{PD}=\overrightarrow{PC}\\ \overrightarrow{AP}+\overrightarrow{PB}=\overrightarrow{PC}-\overrightarrow{PD}

\overrightarrow{AP}+\overrightarrow{PB}=\overrightarrow{PC}+\overrightarrow{DP}              



Since, \overrightarrow{DP}=-\overrightarrow{PD}

\overrightarrow{AP}+\overrightarrow{PB}=\overrightarrow{DP}+\overrightarrow{PC}

By using triangle law in ∆APB, \overrightarrow{AP}+\overrightarrow{PB}=\overrightarrow{AB}

and using triangle law in ∆ DPC, \overrightarrow{DP}+\overrightarrow{PC}=\overrightarrow{DC}

We get

\overrightarrow{AB}=\overrightarrow{DC}            

So, AB is parallel to DC and equal is magnitude.

Hence, ABCD is a parallelogram.

Question 10. Five forces \overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD},\overrightarrow{AE} and  \overrightarrow{AF}act at the vertex of a regular hexagon ABCDEF. Prove that the resultant is 6 \overrightarrow{AO},where o is the centre of hexagon.

Solution:

According to the question,



Prove that

\overrightarrow{AB}+ \overrightarrow{AC}+ \overrightarrow{AD}+\overrightarrow{AE}+ \overrightarrow{AF}=6 \overrightarrow{AO}

Proof:

As we know that the centre(O) of the hexagon bisects the diagonal  \overrightarrow{AD}

So, 

\overrightarrow{AO}=\frac{1}{2} \overrightarrow{AD}; \overrightarrow{BO}=- \overrightarrow{EO}; \overrightarrow{CO}=- \overrightarrow{FO}

Now,

\overrightarrow{AB}+ \overrightarrow{BO}= \overrightarrow{AO}\\  \overrightarrow{AC}+ \overrightarrow{CO}= \overrightarrow{AO}\\   \overrightarrow{AD}+ \overrightarrow{DO}= \overrightarrow{AO}\\    \overrightarrow{AE}+ \overrightarrow{EO}= \overrightarrow{AO}\\     \overrightarrow{AF}+ \overrightarrow{FO}= \overrightarrow{AO}

On adding these equations, we get

( \overrightarrow{AB}+ \overrightarrow{AC}+ \overrightarrow{AD}+ \overrightarrow{AE}+ \overrightarrow{AF})+( \overrightarrow{BO}+ \overrightarrow{CO}+ \overrightarrow{DO}+ \overrightarrow{EO}+ \overrightarrow{FO})=5 \overrightarrow{AO}

( \overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF})+\overrightarrow{DO}=5\overrightarrow{AO}

But  \overrightarrow{DO}=-\overrightarrow{AO}

So, 

\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=6\overrightarrow{AO}

Hence proved

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :