# RD Sharma Class 12 Ex 23.2 Solutions Chapter 23 Algebra of Vectors

Here we provide RD Sharma Class 12 Ex 23.2 Solutions Chapter 23 Algebra of Vectors for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 23.2 Solutions Chapter 23 Algebra of Vectors book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 23.2 Solutions Chapter 23 Algebra of Vectors

### Question 1. If P, Q, and R are three collinear points such that and .  Find the vector Solution:

According to the question, given that

Points P, Q, and R are collinear.

Also, and So,

### Question 2. Given condition that three vectors and form the three sides of a triangle. What are other possibilities?

Solution:

According to the question, given that are three sides of a triangle ABC. [since ] [since ]

So, 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,

or

### Question 3. If and are two non- collinear vectors having the same initial point. What are the vectors represented by and ?

Solution:

According to the question, given that and are two non-collinear vectors having the same initial point.

So, let us considered Now we draw a parallelogram named as ABCD

Using the properties of parallelogram, we get

In ∆ABC,

Using the triangle law, we get …….(i)

In ∆ABD,

Using the triangle law, we get …….(ii)

On solving equation (i) and (ii), we get and are diagonals of a parallelogram whose adjacent sides are and ### Question 4. If is a vector and m is a scalar such that , then what are the alternatives for m and ?

Solution:

According to the question, given that m is a scalar and is a vector such that [since let ]

Now on comparing the coefficients of of LHS and RHS, we get

ma= 0 ⇒ m = 0 or a= 0      …….(i)

mb= 0 ⇒ m = 0 or b= 0          …….(ii)

mc= 0 ⇒ m = 0 or c= 0         …….(iii)

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

m = 0 or a= b= c= 0

m = 0 or m = 0 or ### (iii) Solution:

(i) Let us assume Given that, a = -b

So,

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

a1 = a2         …….(i)

b1 = b2         …….(ii)

c1 = c2         …….(iii)

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

(ii) Given a and b are two vectors such that So, it means the magnitude of vector is equal to the magnitude

of vector , but we cannot find the direction of the vector.

Hence, it is false that

(iii) Given for any vector are equal but we cannot find the direction of the vector of So, it is false.

### Question 6. ABCD is a quadrilateral. Find the sum of the vectors and .

Solution:

According to the question,

ABCD is a quadrilateral.

so,

By using triangle law, we get ……(i)

In ∆ABC,

By using triangle law, we get ……(ii)

Now put the value of in equation (ii), we get

Now on adding on both sides,

### (ii) Solution:

(i) According to the question,

ABCDE is a pentagon,

So,

Using the law of triangle , we get

Using triangle law , , we get

= 0

Hence Proved

(ii) According to the question,

ABCDE is a pentagon,

So,

Using triangle law, , we get

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

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

Hence proved

### Question 9. If P is a point and ABCD is quadrilateral and, show that ABCD is a parallelogram.

Solution:

According to the question

Since, By using triangle law in ∆APB, and using triangle law in ∆ DPC, We get

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

Hence, ABCD is a parallelogram.

### Question 10. Five forces and act at the vertex of a regular hexagon ABCDEF. Prove that the resultant is 6 where o is the centre of hexagon.

Solution:

According to the question,

Prove that

Proof:

As we know that the centre(O) of the hexagon bisects the diagonal So,

Now,

On adding these equations, we get But So,

Hence proved

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