Skip to content
Related Articles

Related Articles

Class 12 RD Sharma Solutions – Chapter 24 Scalar or Dot Product – Exercise 24.2
  • Last Updated : 07 Apr, 2021

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 \vec{a} and \vec{b} 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 \vec{OP}=\frac{2\vec{a}+\vec{b}}{3}

Position vector of Q is \vec{OQ}=\frac{\vec{a}+2\vec{b}}{3}

Now, OP2 + OQ2|\overrightarrow{OP}|^2+|\overrightarrow{OQ}|^2

⇒ (\frac{2\vec{a}+\vec{b}}{3}).(\frac{2\vec{a}+\vec{b}}{3})+(\frac{\vec{a}+2\vec{b}}{3}).(\frac{\vec{a}+2\vec{b}}{3})

⇒ \frac{4|\vec{a}|^2+4\vec{a}.\vec{b}+|\vec{b}|^2+|\vec{a}|^2+4\vec{a}.\vec{b}+4|\vec{b}|^2}{9}

We know that \vec{a}.\vec{b}=0, since \vec{a} and \vec{b} are perpendicular.

⇒ \frac{5}{9}[|\vec{a}|^2+|\vec{b}|^2]



By using Pythagoras theorem, we get

\frac{5}{9}|\overrightarrow{AB}|^2

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 \vec{a}     and \vec{b}     respectively.

Now, Position vector of E is \overrightarrow{OE}=\frac{\vec{a}+\vec{b}}{2}

Using triangle law of vector addition, \overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}

⇒ \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}     

⇒ \overrightarrow{AB}=\vec{b}-\vec{a}

Since the diagonal bisect each other at 90°, \vec{AB}.\vec{OC}=0

⇒ (\vec{b}-\vec{a}).(\vec{b}-\vec{a})=0

⇒ |\vec{b}|^2-|\vec{a}|^2=0     

⇒ |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 AB=\vec{b},AC=\vec{c}

Since AB and AC are perpendicular to each other, \vec{a}.\vec{b}=0

Now, AB2 + AC2|\vec{b}|^2+|\vec{c}|^2     …(1)

From triangle law of vector addition, \overrightarrow{AC}^2=\overrightarrow{AB}^2+\overrightarrow{BC}^2

⇒ \overrightarrow{BC}^2=\overrightarrow{AC}^2-\overrightarrow{AB}^2

⇒ \overrightarrow{BC}^2=\vec{c}^2-\vec{b}^2

Now, BC2|\overrightarrow{BC}|^2=|\vec{c}-\vec{b}|^2

⇒ (\vec{c}-\vec{b}).(\vec{c}-\vec{b})

⇒ |\vec{c}|^2+|\vec{b}|^2-2\vec{c}.\vec{b}     

⇒ |\vec{c}|^2+|\vec{b}|^2      …(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 \vec{b},\vec{d}  respectively.

Using triangle law of vector addition, we have \overrightarrow{AD}+\overrightarrow{DB}=\overrightarrow{AB}

⇒ \overrightarrow{DB}=\overrightarrow{AB}-\overrightarrow{AD}

⇒ \overrightarrow{DB}=\vec{b}-\vec{d}

In triangle ABC, 

\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\vec{b}+\vec{d}

Now, Squares of sides of parallelogram = AB2 + BC2 + CD2 + DA2

⇒ |\overrightarrow{AD}|^2+|\overrightarrow{AB}|^2+|\overrightarrow{-AB}|^2+|-\overrightarrow{AD}|^2

⇒ 2|\overrightarrow{AB}|^2+2|\overrightarrow{AD}|^2

⇒ 2|\vec{b}|^2+2|\vec{d}|^2      …(1)

Also, Squares of diagonals = DB2 + AC2

⇒ |\overrightarrow{DB}|^2+|\overrightarrow{AC}|^2

⇒ |\vec{b}|^2-2\vec{b}.\vec{d}+|\vec{d}|^2+|\vec{b}|^2+2\vec{b}.\vec{d}+|\vec{d}|^2

⇒ 2|\vec{b}|^2+2|\vec{d}|^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 \vec{b},\vec{d}   respectively.

Now, Using triangle law of vector addition, \overrightarrow{PQ}=\overrightarrow{PB}+\overrightarrow{BQ}

⇒ \frac{\overrightarrow{AB}}{2}+\frac{\overrightarrow{BC}}{2}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{BC})

⇒ \overrightarrow{PQ}=\frac{1}{2}(\overrightarrow{AC})

Similarly, 

\overrightarrow{SR}=\overrightarrow{SD}+\overrightarrow{DR}

⇒ \frac{\overrightarrow{AD}}{2}+\frac{\overrightarrow{DC}}{2}=\frac{1}{2}(\overrightarrow{AD}+\overrightarrow{DC})

\overrightarrow{SR}=\frac{1}{2}(\overrightarrow{AC})

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,

⇒ |\overrightarrow{PQ}|^2=\frac{1}{4}|\overrightarrow{AC}|^2     …(1)

Also, from figure, \overrightarrow{PS}=\overrightarrow{AS}-\overrightarrow{PA}

⇒ \frac{\overrightarrow{AD}}{2}-\frac{-\overrightarrow{AB}}{2}=\frac{1}{2}(\overrightarrow{AD}+\overrightarrow{AB})

⇒ \overrightarrow{PS}=\frac{1}{2}(\overrightarrow{AC})

⇒ |\overrightarrow{PS}|^2=\frac{1}{4}(|\overrightarrow{AC}|^2)     …(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 \vec{a},\vec{c}   respectively.

From figure, \overrightarrow{OB} = \overrightarrow{OA}+\overrightarrow{AB}

⇒ \overrightarrow{OB} = \vec{a}+\vec{c}

From figure, \overrightarrow{OC} = \overrightarrow{OA}+\overrightarrow{AC}

⇒ \overrightarrow{AC} = \vec{c}-\vec{a}

Now, \overrightarrow{OB}.\overrightarrow{AC}=(\vec{c}+\vec{a}).(\vec{c}-\vec{a})

⇒ |\vec{c}|^2-|\vec{a}|^2

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

⇒ \overrightarrow{OB}.\overrightarrow{AC}=|\vec{c}|^2-|\vec{a}|^2=0

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 \vec{b},\vec{d}   respectively.

From figure, \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}

⇒ \overrightarrow{AC}=\vec{b}+\vec{d}

Similarly, \vec{AD}=\vec{AB}+\vec{BD}

⇒ \overrightarrow{BD}=\vec{d}-\vec{b}

If the diagonals are perpendicular, then \overrightarrow{AC}.\overrightarrow{BD}=0

⇒ (\vec{b}+\vec{d}).(\vec{b}-\vec{d})=0

⇒ |\vec{b}|^2-|\vec{d}|^2=0

⇒ |\vec{b}|=|\vec{d}|

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 \vec{b},\vec{c}  respectively.

Position vector of AD is \overrightarrow{AD}=\frac{\vec{b}+\vec{c}}{2}

Position vector of CD is \overrightarrow{CD}=\overrightarrow{AD}-\overrightarrow{AC}

⇒ \overrightarrow{CD}=\frac{\vec{b}+\vec{c}}{2}-\vec{c}=\frac{\vec{b}-\vec{c}}{2}

Now, AB2 + AC2|\overrightarrow{AB}|^2+|\overrightarrow{AC}|^2=|\vec{b}|^2+|\vec{c}|^2   …(1)

Also, 2(AD2 + CD2) = 2(|\vec{AD}|^2+|\vec{CD}|^2)

⇒ 2(\frac{|\vec{b}|^2+|\vec{c}|^2+2\vec{b}.\vec{c}}{2}+\frac{|\vec{b}|^2+|\vec{c}|^2-2\vec{b}.\vec{c}}{2})

⇒ |\vec{b}|^2+|\vec{c}|^2  …(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 \vec{b},\vec{c} respectively.

Now, position vector of AD is \overrightarrow{AD}=\frac{\vec{b}+\vec{c}}{2}

Using triangle law of vector addition, \overrightarrow{AC}=\overrightarrow{BC}+\overrightarrow{AB}

⇒ \overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=\vec{c}-\vec{b}

Since, AD and BC are perpendicular, \overrightarrow{BC}.\overrightarrow{AD}=0

⇒ (\vec{c}-\vec{b}).\frac{\vec{b}+\vec{c}}{2}=0

⇒ (\vec{c}-\vec{b}).(\vec{c}+\vec{b})=0

⇒ |\vec{c}|^2-|\vec{b}|^2=0

⇒ |\vec{c}|=|\vec{b}|  

⇒ 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 \vec{b},\vec{c},\vec{d} respectively.

Let P and Q are midpoints of AC, BD.

Position vector of P is \frac{\overrightarrow{AC}}{2}=\frac{\vec{c}}{2}

Position vector of Q is \frac{\vec{b}+\vec{d}}{2}

Now, AB2 + BC2 + CD2 + DA2|\overrightarrow{AB}|^2+|\overrightarrow{BC}|^2+|\overrightarrow{CD}|^2+|\overrightarrow{DA}|^2

⇒ |\vec{b}|^2+|\vec{c}-\vec{b}|^2+|\vec{d}-\vec{c}|^2+|\vec{d}|^2

⇒ 2|\vec{b}|^2+2|\vec{c}|^2+2|\vec{d}|^2-2\vec{b}.\vec{c}-2\vec{c}.\vec{d}   …(1)

Also, AC2 + BD2 + 4PQ2|\overrightarrow{AC}|^2+|\overrightarrow{BD}|^2+4|\overrightarrow{PQ}|^2

⇒ |\vec{c}|^2+|\vec{d}-\vec{b}|^2+4|\frac{|\vec{b}+\vec{d}|^2}{2}-\frac{|\vec{c}|^2}{2}|^2

⇒ |\vec{c}|^2+|\vec{d}-\vec{b}|^2+|\vec{b}+\vec{d}|^2-2|\vec{b}+\vec{d}|.|\vec{c}|+|\vec{c}|^2

⇒ 2|\vec{b}|^2+2|\vec{c}|^2+2|\vec{d}|^2-2\vec{b}.\vec{c}-2\vec{c}.\vec{d}    …(2)

 From eq(1) and eq(2), we get, 

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

Therefore, Hence proved.

My Personal Notes arrow_drop_up
Recommended Articles
Page :