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

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.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter24
Exercise24.3
CategoryRD Sharma Solutions

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 OP2  + OQ= 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, OP+ 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, AB+ 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,

⇒ ABACBC2 

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 = ABBCCDDA2

⇒ |\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 = DBAC2

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

Also, 2(AD+ 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

AB+ AC= 2(AD+ 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 AB+ BC+ CD+ DA= AC+ BD+ 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, AB+ BC+ CD+ 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, 

AB+ BC+ CD+ DA= AC+ BD+ 4PQ2

Therefore, Hence proved.

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