RD Sharma Solutions Class 12 Scalar or Dot Product Exercise 24.2

RD Sharma Solutions for Class 12 Maths Chapter 24 Scalar or Dot Product Exercise 24.2 can be found here. Proper utilization of these solutions will help students achieve excellent marks in their board exams. To accelerate the academic performance of all the students belonging to Class 12, experts at BYJU’S, with vast experience, have formulated the RD Sharma Solutions for Class 12 Maths. Students can now easily download the PDF containing the solutions of this exercise from the link provided below.

RD Sharma Solution for Class 12 Maths Chapter 24 Exercise 2

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Access Answers for Rd Sharma Solution Class 12 Maths Chapter 24 Exercise 2

Exercise 24.2

1. Solution:

RD Sharma Solutions for Class 12 Maths Chapter 24 - 15

∴ OP2 + OQ2 = 5/9 AB2

2. Solution:

Let OACB be a quadrilateral such that its diagonal bisect each other at right angles.

We know that, if the diagonals of a quadrilateral bisect each other, then it’s a parallelogram.

Thus, OACB is a parallelogram

So,

OA = BC and OB = AC

Now,

Taking O as the origin. Let
RD Sharma Solutions for Class 12 Maths Chapter 24 - 16be the position vector of A and B

AB and OC be the diagonals of a quadrilateral, which bisect each other at right angles.

RD Sharma Solutions for Class 12 Maths Chapter 24 - 17
RD Sharma Solutions for Class 12 Maths Chapter 24 - 18

Similarly,

OA = OB = BC = CA

Therefore, OACB is a rhombus.

RD Sharma Solutions for Class 12 Maths Chapter 24 - 193. Solution:

Let ∆AOB be a right-angle triangle with right angle at O.

Required to prove: AB2 = OA2 + OB2

Taking O as the origin, we have

RD Sharma Solutions for Class 12 Maths Chapter 24 - 20

to be the position vector of A and B, respectively.

Now, as OB is perpendicular to OA, their dot product equals zero

So, we have

RD Sharma Solutions for Class 12 Maths Chapter 24 - 21

And,

RD Sharma Solutions for Class 12 Maths Chapter 24 - 22

Therefore,

AB2 = OA2 + OB2

  • Hence proved

RD Sharma Solutions for Class 12 Maths Chapter 24 - 234. Solution:

Let OAC be a right triangle, right-angled at O.

Now, taking O as the origin

Let
RD Sharma Solutions for Class 12 Maths Chapter 24 - 24be the position vector of
RD Sharma Solutions for Class 12 Maths Chapter 24 - 25.

RD Sharma Solutions for Class 12 Maths Chapter 24 - 26

  • Hence proved

5. Solution:

RD Sharma Solutions for Class 12 Maths Chapter 24 - 27

Given, ABCD is a rectangle

Let P, Q, R and S be the midpoints of the sides AB, BC, CD and DA, respectively.

Now,

RD Sharma Solutions for Class 12 Maths Chapter 24 - 28

From (iii) and (iv), we get

(PQ)2 = (PS)2

⇒ PQ = PS

So, the adjacent sides of PQRS are equal

Hence, PQRS is a rhombus.

6. Solution:

RD Sharma Solutions for Class 12 Maths Chapter 24 - 29

Let OABC be a rhombus whose diagonals OB and AC intersect at point D

And let O be the origin

Let the position vector of A and C be
RD Sharma Solutions for Class 12 Maths Chapter 24 - 30respectively then,

RD Sharma Solutions for Class 12 Maths Chapter 24 - 31

7. Solution:

RD Sharma Solutions for Class 12 Maths Chapter 24 - 32

Let ABCD be a rectangle

Taking A as the origin, we have position vectors of points B and D to be
RD Sharma Solutions for Class 12 Maths Chapter 24 - 33respectively

By parallelogram law,

RD Sharma Solutions for Class 12 Maths Chapter 24 - 34

Hence, ABCD is a square.

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