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:

∴ 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
be the position vector of A and B

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

Similarly,

OA = OB = BC = CA

Therefore, OACB is a rhombus.

3. 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

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

And,

Therefore,

AB² = OA² + OB²

• Hence proved

4. Solution:

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

Now, taking O as the origin

Let
be the position vector of
.

• Hence proved

5. Solution:

Given, ABCD is a rectangle

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

Now,

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:

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
respectively then,

7. Solution:

Let ABCD be a rectangle

Taking A as the origin, we have position vectors of points B and D to be
respectively

By parallelogram law,

Hence, ABCD is a square.