# Class 12 RD Sharma Solutions – Chapter 24 Scalar or Dot Product – Exercise 24.2

### 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 = OBTherefore, 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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