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 OP²  + OQ² = 5/9 AB².

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, OP² + OQ² = 

⇒ 

⇒ 

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, AB² + AC² =  …(1)

From triangle law of vector addition, 

⇒ 

⇒ 

Now, BC² = 

⇒ 

⇒  

⇒   …(2)

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

⇒ AB² + AC² = BC² 

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 = AB² + BC² + CD² + DA²

⇒ 

⇒ 

⇒    …(1)

Also, Squares of diagonals = DB² + AC²

⇒ 

⇒ 

⇒    …(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 AB² + AC² = 2(AD² + CD²).

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, AB² + AC² =  …(1)

Also, 2(AD² + CD²) = 

⇒ 

⇒  …(2)

From eq(1) and eq(2), we get

AB² + AC² = 2(AD² + CD²)

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 AB² + BC² + CD² + DA² = AC² + BD² + 4PQ², 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, AB² + BC² + CD² + DA² = 

⇒ 

⇒   …(1)

Also, AC² + BD² + 4PQ² = 

⇒ 

⇒ 

⇒    …(2)

 From eq(1) and eq(2), we get, 

AB² + BC² + CD² + DA² = AC² + BD² + 4PQ²

Therefore, Hence proved.