Here we provide RD Sharma Class 12 Ex 24.3 Solutions Chapter 24 Scalar or Dot Product for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 24.3 Solutions Chapter 24 Scalar or Dot Product book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 24 |

Exercise | 24.3 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Class 12 Ex 24.3 Solutions Chapter 24 Scalar or Dot Product**

**Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed **

### Question 1. In a triangle OAB, if P, Q are points of trisection of AB, Prove that OP^{2} + OQ^{2 }= 5/9 AB^{2}.

**Solution:**

Given that in triangle OAB,

∠AOB = 90°, and P, Q are points of trisection of AB.

Consider ‘O’ as origin, the position vectors of A and B are and respectively.

Since P and Q are points of trisection of AB, AP:PB = 1:2 and AQ:QB = 2:1.

From section formula position vector of P is

Position vector of Q is

Now, OP^{2 }+ OQ^{2} =

⇒

⇒

We know that , since and are perpendicular.

⇒

By using Pythagoras theorem, we get

Hence proved.

### Question 2. Prove that if the diagonals of a quadrilateral bisects each other at right angles, then it is a rhombus.

**Solution:**

Let us considered OABC be quadrilateral and the diagonals AB and OC bisect each other at 90°.

Consider ‘0’ as origin and the position vectors of A and B are given by and respectively.

Now, Position vector of E is

Using triangle law of vector addition,

⇒

⇒

Since the diagonal bisect each other at 90°,

⇒

⇒

⇒ |a| = |b|

⇒** **OA = OB

Therefore, we proved that adjacent sides of quadrilateral are equal, if its diagonals bisect each other at 90°.

### Question 3. Prove by vector method that in a right-angled triangle, the square of the hypotenuse is equal to sum of squares of other two sides(Pythagoras Theorem).

**Solution:**

Let us considered ABC be the right angled triangle with ∠BAC = 90°.

Consider ‘A’ as origin and the position vectors

Since AB and AC are perpendicular to each other,

Now, AB^{2 }+ AC^{2} = …(1)

From triangle law of vector addition,

⇒

⇒

Now, BC^{2} =

⇒

⇒

⇒ …(2)

Therefore, from eq(1) and eq(2) we get,

⇒ *AB*^{2 }+ *AC*^{2 }= *BC*^{2}

Hence proved.

### Question 4. Prove by vector method that the sum of squares of diagonals of the parallelogram is equal to sum of squares of its sides.

**Solution:**

Let us considered ABCD be a parallelogram and AC, BD are its diagonals.

Consider A as origin, Let the position vectors of AB, AD are respectively.

Using triangle law of vector addition, we have

⇒

⇒

In triangle ABC,

Now, Squares of sides of parallelogram = *AB*^{2 }+ *BC*^{2 }+ *CD*^{2 }+ *DA*^{2}

⇒

⇒

⇒ …(1)

Also, Squares of diagonals = *DB*^{2 }+ *AC*^{2}

⇒

⇒

⇒ …(2)

By, observing eq(1) and eq(2),

We proved that sum of squares of sides of a parallelogram is equal to sum of squares of its diagonals.

### Question 5. Prove using vector method that the quadrilateral obtained by joining the mid-points of adjacent sides of the rectangle is a rhombus.

**Solution: **

Let us considered ABCD is a rectangle and P, Q, R, S are midpoints of AB, BC, CD, DA respectively.

Consider A as origin, the position vectors of AB, AD are respectively.

Now, Using triangle law of vector addition,

⇒

⇒

Similarly,

⇒

⇒

By observing, we find that PQ || SR, so we can say it is a parallelogram

Let us find if it forms a rhombus by calculating length of adjacent sides,

⇒ …(1)

Also, from figure,

⇒

⇒

⇒ …(2)

From eq(1) and eq(2) PQ and PS are adjacent sides and |PQ|=|PS|,

so PQRS is a Rhombus.

Therefore, Hence proved

### Question 6. Prove that diagonals of rhombus are perpendicular bisectors of each other.

**Solution:**

Let us considered OABC be a Rhombus, OB and AC are diagonals of Rhombus.

Consider O as origin, position vectors of OA and OC are respectively.

From figure,

⇒

From figure,

⇒

Now,

⇒

We know, that adjacent sides are equal in a Rhombus,

⇒

Therefore, diagonals of rhombus are perpendicular bisectors of each other.

### Question 7. Prove that diagonals are of the rectangle are perpendicular if and only if the rectangle is a square.

**Solution:**

Let us considered ABCD is a rectangle, AC, BD are diagonals of rectangle.

Consider A as origin, Position vectors of AB, AD are respectively.

From figure,

⇒

Similarly,

⇒

If the diagonals are perpendicular, then

⇒

⇒

⇒

Therefore, If the diagonals of rectangle are perpendicular, Then it is a square.

### Question 8. If AD is median of triangle ABC, using vectors prove that AB^{2 }+ AC^{2 }= 2(AD^{2 }+ CD^{2}).

**Solution:**

Let us considered ABC is a triangle and AD is median.

Consider A is a origin, position vectors of AB and AC are respectively.

Position vector of AD is

Position vector of CD is

⇒

Now, AB^{2 }+ AC^{2} = …(1)

Also, 2(AD^{2 }+ CD^{2}) =

⇒

⇒ …(2)

From eq(1) and eq(2), we get

AB^{2 }+ AC^{2 }= 2(AD^{2 }+ CD^{2})

Therefore, Hence proved.

### Question 9. If the median to the base of triangle is perpendicular to base, then triangle is isosceles.

**Solution:**

Let us considered ABC be a triangle and AD is median.

Consider A as origin, position vector of AB, AC are respectively.

Now, position vector of AD is

Using triangle law of vector addition,

⇒

Since, AD and BC are perpendicular,

⇒

⇒

⇒

⇒

⇒ AC = AB

Therefore, Triangle ABC is an Isosceles triangle.

### Question 10. In a quadrilateral ABCD, prove that AB^{2 }+ BC^{2 }+ CD^{2 }+ DA^{2 }= AC^{2 }+ BD^{2 }+ 4PQ^{2}, where P, Q are midpoints of diagonals AC and BD.

**Solution:**

Let us considered ABCD be a quadrilateral, AC, BD are diagonals.

Consider A as origin, the position vectors of AB, AC, AD are respectively.

Let P and Q are midpoints of AC, BD.

Position vector of P is

Position vector of Q is

Now, AB^{2 }+ BC^{2 }+ CD^{2 }+ DA^{2} =

⇒

⇒ …(1)

Also, AC^{2} + BD^{2} + 4PQ^{2} =

⇒

⇒

⇒ …(2)

From eq(1) and eq(2), we get,

AB^{2 }+ BC^{2 }+ CD^{2 }+ DA^{2 }= AC^{2 }+ BD^{2 }+ 4PQ^{2}

Therefore, Hence proved.

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