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

**12th Maths 6th Chapter Solutions 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)

**12th Maths Exercise 6.1 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.

**12th Maths 6th Chapter 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.

This gives ∠APB = 90°. Hence the result.

**12th Maths Chapter 6 Exercise 6.1 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 Guru 12th Maths 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.

\(\overrightarrow{\mathrm{AB}}\) ⊥^{r} to \(\overrightarrow{\mathrm{BC}}\)

⇒ ABCD is a rectangle.

**Ex 6.1 Class 12 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}

**6.1 12th Maths 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’

**12th State Board Maths Solution Book Pdf English Medium 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:

Similarly we can prove

Area of ∆GBC = Area of ∆GAC = \(\frac{1}{3}\) [Area of ∆ABC]

**12th Maths Chapter 6 Solution Question 9.**

Using vector method, prove that cos(α – β) = cos α cos β + sin α sin β.

Solution:

From (1) and (2), we get

cos(α + β) = cos α cos β + sin α sin β

**Samacheer Kalvi 12 Maths Solutions 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

From (1) & (2), we get

sin (α + β) = sin α cos β + cos α sin β

**12 Maths Samacheer Kalvi Solutions 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:

From (1) & (2), we get

Work done by the force = \(\overrightarrow{\mathrm{F}} \cdot \vec{d}\) = 56 + 8 + 16 = 80 units.

**Std 12 Maths Solutions Samacheer Kalvi 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:

**Class 12 Maths Solutions Samacheer 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 Guru 12 Maths 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 Additional Problems

**Samacheer Kalvi Class 12 Maths Solutions 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:

**12 Maths Solutions Samacheer Kalvi Question 2.**

If the position vectors of three points A, B and C are respectively and \(7(\vec{i}+\vec{k})\). Find \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}\). Interpret the result geometrically.

Solution:

**Samacheer Kalvi.Guru 12th Maths 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:

**Maths 12th Guide Question 4.**

Show that the area of a parallelogram having diagonals is \(5 \sqrt{3}\)

Solution: