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

In Chapter 23 Algebra of Vectors of RD Sharma's Class 12 mathematics textbook the focus is on understanding and applying vector algebra. This chapter covers various operations and properties of vectors in which are fundamental in both mathematics and physics. Exercise 23.2 specifically deals with the practical problems that require the application of vector addition, subtraction, and scalar multiplication. These concepts are crucial for the solving problems involving forces, velocities, and other vector quantities in different contexts.

Algebra of Vectors

The Algebra of Vectors involves the study of operations such as the addition, subtraction, and scalar multiplication of the vectors. It provides the tools to work with the vector quantities in a structured manner. The Vector operations are essential for solving real-world problems in physics and engineering where quantities have both magnitude and direction. Understanding vector algebra helps in solving problems related to forces, motion, and various physical phenomena making it a vital area of study in mathematics.

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}   \overrightarrow{OD}=-\overrightarrow{OH}        .......(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

Read more :

• Vector Algebra
• Section Formula – Vector Algebra

Summary

Exercise 23.2 in RD Sharma's Class 12 textbook typically focuses on vector operations, particularly addition, subtraction, and scalar multiplication. This exercise introduces students to the fundamental properties of vector algebra, including the parallelogram law of vector addition, the triangle law, and the concept of unit vectors. Students learn to perform vector calculations in both two and three dimensions, often using component form (i, j, k notation). The exercise emphasizes understanding vector equality, magnitude calculation, and the geometric interpretation of vector operations. It also covers topics such as finding the direction cosines of vectors and solving problems involving position vectors, laying a crucial foundation for more advanced vector concepts.

Practice Questions

1).Given vectors a = 3i - 2j + k and b = i + 4j - 2k, find 2a - 3b.

2).If |a| = 3 and |b| = 4, and the angle between them is 60°, find |a + b|.

3).Prove that the diagonals of a parallelogram bisect each other.

4).Find a unit vector in the direction of v = 6i - 8j + 10k.

5).If a + b = 2i + 3j and a - b = 4i - j, find vectors a and b.

6).Prove that (a + b) + c = a + (b + c) for any three vectors a, b, and c.

7).Find the value of λ if vectors 2i + λj - k and i - 3j + 2k are parallel.

8).If a = 2i - j + 2k and b = i + 2j - k, find the vector 3a - 2b + 5(a + b).

9).Determine the direction cosines of the vector 3i + 4j - 12k.

10).Prove that the medians of a triangle divide each other in the ratio 2:1.