Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1

Question 1.
Prove by vector method that if a line is drawn from the centre of a circle of a circle to the midpoint of a chord, then the line is perpendicular to the chord.
Solution:
Let ‘C’ be the mid point of the chord AB
Take ‘O’ on the centre of the circle.
Since, OA = OB (Radii)
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 1
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 2

Question 2.
Prove by vector method that the median to the base of an isosceles triangle is perpendicular to the base.
Solution:
Let OAB be an isosceles triangle with OA = OB
Let OC be the median to the base AB
C is the midpoint of AB
Take O as origin.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 3

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1

Question 3.
Prove by vector method that an angle in a semi-circle is a right angle.
Solution:
Let AB be the diameter of the circle with centre ‘O’
Let P be any point on the semi-circle.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 4
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 5
This gives ∠APB = 90°. Hence the result.

Question 4.
Prove by vector method that the diagonals of a rhombus bisect each other at right angles.
Solution:
Let ABCD be a rhombus
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 6

Question 5.
Using vector method, prove that if the diagonals of a parallelogram are equal, then it is a rectangle.
Solution:
Let ABCD be a parallelogram
To prove ABCD be a rectangle provided the diagonals are equal.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 7
\(\overrightarrow{\mathrm{AB}}\) ⊥r to \(\overrightarrow{\mathrm{BC}}\)
⇒ ABCD is a rectangle.

Question 6.
Prove by vector method that the area of the quadrilateral ABCD having diagonals AC and BD is \(\frac{1}{2}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}|\)
Solution:
Vector area of quadrilateral ABCD = {Vector area of ∆ABC} + {Vector area of ∆ ACD}
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 8

Question 7.
Prove by vector method that the parallelogram on the same base and between the same parallels are equal in area.
Solution:
Let ABCD and ABC’D’ be two parallelogram between the parallels with same base
To prove: Area of ABCD = Area of ABC’D’
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 9
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 10

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1

Question 8.
If G is the centroid of a AABC, prove that.
(area of ∆GAB) = (area of ∆GBC) = (area of ∆GAC) = \(\frac{1}{3}\) [area of ∆ABC]
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 11
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 12
Similarly we can prove
Area of ∆GBC = Area of ∆GAC = \(\frac{1}{3}\) [Area of ∆ABC]

Question 9.
Using vector method, prove that cos(α – β) = cos α cos β + sin α sin β.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 13
From (1) and (2), we get
cos(α + β) = cos α cos β + sin α sin β

Question 10.
Prove by vector method that sin(α + β) = sin α cos β + cos α sin β.
Solution:
Take two points A and B on the unit circle with centre as origin ‘O’, so \(|\overrightarrow{\mathrm{OA}}|=|\overrightarrow{\mathrm{OB}}|\) = 1
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 14
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 15
From (1) & (2), we get
sin (α + β) = sin α cos β + cos α sin β

Question 11.
A particle acted on by constant forces \(8 \vec{i}+2 \vec{j}-6 \vec{k}\) and \(\overrightarrow{6 i}+2 \vec{j}-2 \vec{k}\) is displaced from the point (1, 2, 3) to the point (5, 4, 1). Find the total work done by the forces.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 16
From (1) & (2), we get
Work done by the force = \(\overrightarrow{\mathrm{F}} \cdot \vec{d}\) = 56 + 8 + 16 = 80 units.

Question 12.
Forces of magnitude \(5 \sqrt{2}\) and \(10 \sqrt{2}\) units acting in the directions \((3 \vec{i}+4 \vec{j}+5 \vec{k})\) and \((10 \vec{i}+6 \vec{j}-8 \vec{k})\), respectively, act on a particle which is displaced from the point with position vector \((4 \vec{i}-3 \vec{j}-2 \vec{k})\) to the point with position vector \((\overrightarrow{6 i}+\vec{j}-3 \vec{k})\). Find the work done by the forces.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 17
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 18

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1

Question 13.
Find the magnitude and direction cosines of the torque of a force represented by \(3 \vec{i}+4 \vec{j}-5 \vec{k}\) about the point with position vector \(2 \vec{i}-3 \vec{j}+4 \vec{k}\) acting through a point whose position vector is \(\overrightarrow{4 i}+2 \vec{j}-3 \vec{k}\).
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 19

Question 14.
Find the torque of the resultant of the three forces represented by \(-3 \vec{i}+6 \vec{j}-3 \vec{k}\), \(\overrightarrow{4 i}-10 \vec{j}+12 \vec{k}\) and \(\overrightarrow{4 i}+7 \vec{j}\) acting at the point with position vector \(8 \vec{i}-\overrightarrow{6} \vec{j}-4 \vec{k}\), about the point with position vector \(18 \vec{i}+3 \vec{j}-9 \vec{k}\)
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 20
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 21

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 Additional Problems

Question 1.
The work done by the force \(\overrightarrow{\mathrm{F}}=a \vec{i}+\vec{j}+\vec{k}\) in moving the point of application from (1, 1, 1) to (2, 2, 2) along a straight line is given to be 5 units. Find the value of a.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 22

Question 2.
If the position vectors of three points A, B and C are respectively Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 23 and \(7(\vec{i}+\vec{k})\). Find \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}\). Interpret the result geometrically.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 24

Question 3.
A force given by and \(3 \vec{i}+2 \vec{j}-4 \vec{k}\) is applied at the point (1, -1, 2). Find the moment of the force about the point (2, -1, 3).
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 25

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1

Question 4.
Show that the area of a parallelogram having diagonals Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 26 is \(5 \sqrt{3}\)
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 27

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