Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.1 Textbook Questions and Answers, Notes.

## 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 to the midpoint of a chord; then the line is perpendicular to the chord.
Solution:
A circle with centre at O. AB is chord of the circle and OP bisects AB (ie) AP = PB
To prove $$\overline { OP }$$ ⊥ $$\overline { AB }$$ O is the position vector
∴ $$\overline { OA }$$ = $$\overline { OB }$$ = Radius
Position vector of P

Hence proved

Question 2.
Prove by vector method that the median to the base of an isosceles triangle is perpendicular to the base.
Solution:
In isosceles ΔABC
Let AB = AC and AD is the median
D is the mid point of BC

$$\overline { DA }$$.$$\overline { DB }$$ = 0
$$\overline { DA }$$ ⊥ $$\overline { DB }$$

Question 3.
Prove by vector method that an angle in a semi-circle is a right angle.
Solution:

Let us consider a circle with centre O and diameter AB.
Let P be any point on the semi-circle.
Let us prove that ∠APB = 90°
W.K.T OA = OB = OP (∵ radius)
$$\overline { PA }$$ = $$\overline { PO }$$ + $$\overline { OA }$$
$$\overline { PB }$$ = $$\overline { PO }$$ + $$\overline { OB }$$
= $$\overline { PO }$$ – $$\overline { OA }$$
$$\overline { PA }$$. $$\overline { PB }$$ = ($$\overline { PO }$$ + $$\overline { OA }$$)($$\overline { PO }$$ – $$\overline { OA }$$)
= $$\overline { PO }$$² – $$\overline { OA }$$² = 0
$$\overline { PA }$$ ⊥ $$\overline { PB }$$
⇒ ∠APB = 90°. Hence proved.

Question 4.
Prove by vector method that the diagonals of a rhombus bisect each other at right angles.
Solution:
ABCD is a rhombus

$$\overline { AB }$$ = $$\overline { a }$$ and $$\overline { AD }$$ = $$\overline { b }$$
Here AB = BC = CD = DA
(ie) |$$\overline { a }$$| = |$$\overline { b }$$|
$$\overline { AC }$$ = $$\overline { AB }$$ + $$\overline { BC }$$
= $$\overline { a }$$ + $$\overline { b }$$
$$\overline { BD }$$ = $$\overline { BC }$$ + $$\overline { CD }$$
= $$\overline { AD }$$ – $$\overline { AB }$$
= $$\overline { b }$$ – $$\overline { a }$$
$$\overline { AC }$$.$$\overline { BD }$$ = ($$\overline { a }$$ + $$\overline { b }$$).($$\overline { b }$$ – $$\overline { a }$$)
= ($$\overline { b }$$ + $$\overline { a }$$).($$\overline { b }$$ – $$\overline { a }$$)
= ($$\overline { b }$$)² – ($$\overline { a }$$)² = 0 (∵ |$$\overline { a }$$| = |$$\overline { b }$$|)
$$\overline { AC }$$.$$\overline { BD }$$ = 0 ⇒ $$\overline { AC }$$ ⊥ $$\overline { BD }$$
Hence proved.

Question 5.
Using vector method, prove that if the diagonals of a parallelogram are equal, then it is a rectangle.
Solution:

ABCD is a parallelogram with sides
$$\overline { AB }$$ = $$\overline { a }$$, $$\overline { AD }$$ = $$\overline { b }$$ and the diagonals are $$\overline { AC }$$ and $$\overline { BD }$$
$$\overline { AC }$$ = $$\overline { AB }$$ + $$\overline { BC }$$ = $$\overline { AB }$$ + $$\overline { AD }$$ = $$\overline { a }$$ + $$\overline { b }$$
$$\overline { BD }$$ = $$\overline { BA }$$ + $$\overline { AD }$$ = $$\overline { AD }$$ – $$\overline { AB }$$ = $$\overline { b }$$ – $$\overline { a }$$
Since the diagonals are equal
|$$\overline { AC }$$| = |$$\overline { BD }$$|
|$$\overline { a }$$ + $$\overline { b }$$| = |$$\overline { b }$$ – $$\overline { a }$$|
($$\overline { a }$$ + $$\overline { b }$$)² = ($$\overline { b }$$ – $$\overline { a }$$)²
$$\overline { a }$$² + 2$$\overline { a }$$.$$\overline { b }$$ + $$\overline { b }$$² = $$\overline { b }$$² – 2$$\overline { b }$$ $$\overline { a }$$ + $$\overline { a }$$²
4$$\overline { a }$$ $$\overline { b }$$ = 0
$$\overline { a }$$.$$\overline { b }$$ = 0
$$\overline { a }$$ ⊥$$\overline { b }$$ ⇒ ABCD is a rectangle. Since ∠A = 90°.

Question 6.
Prove by vector method that the area of the quadrilateral ABCD having diagonals AC and BD is $$\frac { 1 }{ 2 }$$ |$$\overline { AC }$$| = |$$\overline { BD }$$|
Solution:
Vector area of a quadrilateral ABCD
= Vector area of ΔABC + Vector area of ΔACD

= $$\frac { 1 }{ 2 }$$ |$$\overline { AC }$$| = |$$\overline { BD }$$|

Question 7.
Prove by vector method that the parallelograms on the same base and between the same parallels are equal in area.
Solution:
$$\overline { AB }$$ = $$\overline { a }$$, $$\overline { AD }$$ = $$\overline { b }$$

Vector area of the parallelogram is $$\overline { b }$$ × $$\overline { a }$$ ……(1)
Consider the parallelogram ABB’A’
$$\overline { AB }$$ = $$\overline { a }$$, $$\overline { AB }$$ = m$$\overline { a }$$
Because $$\overline { A’B }$$ is parallel to $$\overline { AB }$$
By law of vectors AA’ = m$$\overline { a }$$ + $$\overline { b }$$
Hence the vector area of parallelogram
ABB’A = $$\overline { a }$$ × (m$$\overline { a }$$ + $$\overline { b }$$)
= m($$\overline { a }$$ × $$\overline { a }$$) + ($$\overline { a }$$ × $$\overline { b }$$)
= 0 + ($$\overline { a }$$ × $$\overline { b }$$)
= $$\overline { a }$$ × $$\overline { b }$$ ……..(2)
By (1) and (2) Area of ABCD = Area of ABB’A’
Hence proved.

Question 8.
If G is the centroid of a ΔABC, prove that (area of ΔGAB) = (area of ΔGBC) = (area of ΔGCA) = $$\frac { 1 }{ 3 }$$ (area of ΔABC).
Solution:
W.K.T the median of a triangle divides it into two triangles of equal area.

In ΔABC, AD is the median
area (ΔABD) = area (ΔACD) ………. (1)
In ΔGBC, GD is die median
area (ΔGBD) = area (ΔGCD) ………. (2)
Sub (2) from (1) we get
area (ΔABD) – area (ΔGBD)
= area (ΔACD) – area (ΔGCD)
area(ΔAGB) = area(ΔAGC) ………… (3)
Similarly
area (ΔAGB) = area (ΔBGC) …………(4)
From (3) and (4) we get
area (ΔAGB) = area(ΔAGC) = area(ΔBGC) ……….. (5)
Now
area(ΔAGB) + area(ΔAGC) + area(ΔBGC) = area(ΔABC)
⇒ area(ΔAGB) + area(ΔAGB) + area(ΔAGB)
= area (ΔABC) (using 5)
⇒ 3area(ΔAGB) = area(ΔABC)
⇒ axea(ΔAGB) = $$\frac { 1 }{ 3 }$$ area(ΔABC) ………..(6)
From (5) and (6) we get
area(ΔAGB) = area(ΔAGB) = area(ΔBGC)
= $$\frac { 1 }{ 3 }$$ area(ΔABC)

Question 9.
Using vector method, prove that
cos(α – ß) = cos α cos ß + sin α sin ß.
Solution:
Let $$\overline { a }$$ = $$\overline { OA }$$, $$\overline { b }$$ = $$\overline { OB }$$ be the unit vectors and which make angles α and ß respectively with positive x-axis where A and B are as in diagram.

Draw AL and BM perpendicular to the X axis, then
$$\overline { OL }$$ = $$\overline { OA }$$ = cos α
|$$\overline { OL }$$| = |$$\overline { OA }$$| cos α = cos α
|$$\overline { LA }$$| = |$$\overline { OA }$$| sin α = sin α
$$\overline { OL }$$ = |$$\overline { OL }$$|$$\hat { i }$$ = cos α $$\hat { i }$$
$$\overline { LA }$$ = sin α (+$$\hat { j }$$)
$$\overline { a }$$ = $$\overline { OA }$$ = $$\overline { OL }$$ + $$\overline { LA }$$
= cosα $$\hat { i }$$ + sinα $$\hat { j }$$ ……… (1)
Similarly $$\overline { b }$$ = cos ß $$\hat { i }$$ + sin ß $$\hat { j }$$ ……… (2)
The angle between $$\overline { a }$$ and $$\overline { b }$$ is α – ß and so $$\overline { a }$$.$$\overline { b }$$ = |$$\overline { a }$$||$$\overline { b }$$| cos(α – ß) = cos(α – ß) ……..(3)
From (1) and (2)
$$\overline { a }$$.$$\overline { b }$$ = (cosα$$\hat { i }$$ + sinα $$\hat { j }$$).(cosß$$\hat { i }$$ + sinß$$\hat { j }$$)
= cos α cos ß + sin α sin ß ………. (4)
From (3) and (4)
cos(α – ß) = cos α cos ß + sin α sin ß

Question 10.
Prove by vector method that
sin(α + ß) = sin α cos ß +cos α sin ß.
Solution:
Let $$\overline { a }$$ = $$\overline { OA }$$, $$\overline { b }$$ = $$\overline { OB }$$ be the unit vectors making angles α and ß respectively with positive x axis where A and B are as shown in the diagram

Draw AL and BM perpendicular to the X axis, then
$$\overline { OL }$$ = $$\overline { OA }$$ cos α
|$$\overline { OL }$$| = |$$\overline { OA }$$| cos α = cos α
|$$\overline { LA }$$| = |$$\overline { OA }$$| sin α = sin α
|$$\overline { OL }$$| = |$$\overline { OL }$$| $$\hat { i }$$ = cos α $$\hat { i }$$
$$\overline { LA }$$ = sin α(-$$\hat { j }$$)
$$\overline { a }$$ = $$\overline { OA }$$ = $$\overline { OL }$$ + $$\overline { LA }$$
= cos α $$\hat { i }$$ + sin α $$\hat { j }$$ ………… (1)
similarly $$\overline { b }$$ = cos ß $$\hat { i }$$ – sin ß $$\hat { j }$$ ………… (2)
The angle between $$\overline { a }$$ and $$\overline { b }$$ is α + ß and the vectors $$\overline { b }$$, $$\overline { a }$$, $$\overline { k }$$ from a right handed system.
$$\overline { b }$$ × $$\overline { a }$$ = |$$\overline { b }$$||$$\overline { a }$$| sin(α + ß)$$\hat { k }$$ = sin(α + ß)$$\hat { k }$$ ………..(1)
$$\overline { b }$$ × $$\overline { a }$$ = $$\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \cos \beta & -\sin \beta & 0 \\ \cos \alpha & \sin \alpha & 0 \end{array}\right|$$
= (sin α cos ß + cos α sin ß)$$\hat { k }$$ ……… (2)
From (1) & (2)
sin(α + ß) = sin α cos ß + cos α sin ß

Question 11.
A particle acted on by constant forces 8$$\hat { i }$$ + 2$$\hat { j }$$ – 6$$\hat { k }$$ and 6$$\hat { i }$$ + 2$$\hat { j }$$ – 2$$\hat { k }$$ is displaced from the point (1, 2, 3) to the point (5, 4, 1). Find the total work done by the forces.
Solution:
$$\overline { OA }$$ = $$\hat { i }$$ + 2$$\hat { j }$$ + 3$$\hat { k }$$
$$\overline { OB }$$ = 5$$\hat { i }$$ + 4$$\hat { j }$$ + $$\hat { k }$$
$$\overline { d }$$ = $$\overline { AB }$$ = $$\overline { OB }$$ – $$\overline { OA }$$
= 4$$\hat { i }$$ + 2$$\hat { j }$$ – 2$$\hat { k }$$
$$\overline { F_1 }$$ = 8$$\hat { i }$$ + 2$$\hat { j }$$ – 6$$\hat { k }$$
and $$\overline { F_2 }$$ = 6$$\hat { i }$$ + 2$$\hat { j }$$ – 2$$\hat { k }$$
$$\overline { F }$$ = $$\overline { F_1 }$$ + $$\overline { F_2 }$$ = 14$$\hat { i }$$ + 4$$\hat { j }$$ – 8$$\hat { k }$$
Work done = $$\overline { F }$$ $$\overline { d }$$
= (14$$\hat { i }$$ + 4$$\hat { j }$$ – 8$$\hat { k }$$).(4$$\hat { i }$$ + 2$$\hat { j }$$ – 2$$\hat { k }$$)
= 56 + 8 + 16
= 80 units

Question 12.
Forces of magnitudes 5√2 and 10√2 units acting in the directions 3$$\hat { i }$$ + 4$$\hat { j }$$ + 5$$\hat { k }$$ and 10$$\hat { i }$$ + 6$$\hat { j }$$ – 8$$\hat { k }$$, respectively, act on a particle which is displaced from the point with position vector 4$$\hat { i }$$ – 3$$\hat { j }$$ – 2$$\hat { k }$$ to the point with position vector 6$$\hat { i }$$ – $$\hat { j }$$ – 3$$\hat { k }$$. Find the work done by the forces.

Resultant force $$\overline { F }$$ = $$\overline { F_1 }$$ + $$\overline { F_2 }$$
= 5√2$$\overline { F_1 }$$ + 10√2 $$\overline { F_2 }$$
= 3$$\hat { i }$$ + 4$$\hat { j }$$ + 5$$\hat { k }$$ + 10$$\hat { i }$$ + 6$$\hat { j }$$ – 8$$\hat { k }$$
= 13$$\hat { i }$$ + 10$$\hat { j }$$ – 3$$\hat { k }$$
$$\overline { OA }$$ = 4$$\hat { i }$$ – 3$$\hat { j }$$ – 2$$\hat { k }$$
$$\overline { OB }$$ = 6$$\hat { i }$$ + $$\hat { j }$$ – 3$$\hat { k }$$
$$\overline { d }$$ = $$\overline { AB }$$ = $$\overline { OB }$$ – $$\overline { OA }$$ = 2$$\hat { i }$$ + 4$$\hat { j }$$ – $$\hat { k }$$
$$\overline { F }$$.$$\overline { d }$$ = (13$$\hat { i }$$ + 10$$\hat { j }$$ – 3$$\hat { k }$$).(2$$\hat { i }$$ + 4$$\hat { j }$$ – $$\hat { k }$$)
= 26 + 40 + 3
= 69 units

Question 13.
13. Find the magnitude and direction cosines of the torque of a force represented by 3$$\hat { i }$$ + 4$$\hat { j }$$ – 5$$\hat { k }$$ about the point with position vector 2$$\hat { i }$$ – 3$$\hat { j }$$ + 4$$\hat { k }$$ acting through a point whose position vector is 4$$\hat { i }$$ + 2$$\hat { j }$$ – 3$$\hat { k }$$
Solution:
$$\overline { OA }$$ = 2$$\hat { i }$$ – 3$$\hat { j }$$ + 4$$\hat { k }$$
$$\overline { OB }$$ = 4$$\hat { i }$$ + 2$$\hat { j }$$ – 3$$\hat { k }$$
$$\hat { r }$$ = $$\overline { AB }$$ = $$\overline { OB }$$ – $$\overline { OA }$$
= 2$$\hat { i }$$ + 5$$\hat { j }$$ – 7$$\hat { k }$$
= 3$$\hat { i }$$ + 4$$\hat { j }$$ – 5$$\hat { k }$$

Question 14.
Find the torque of the resultant of the three forces represented by -3$$\hat { i }$$ + 6$$\hat { j }$$ – 3$$\hat { k }$$, 4$$\hat { i }$$ – 10$$\hat { j }$$ + 12$$\hat { k }$$ and 4$$\hat { i }$$ + 7$$\hat { j }$$ acting at the point with position vector 8$$\hat { i }$$ – 6$$\hat { j }$$ – 4$$\hat { k }$$ about the point with position vector 18$$\hat { i }$$ + 3$$\hat { j }$$ – 9$$\hat { k }$$
Solution:
$$\overline { F_1 }$$ = -3$$\hat { i }$$ + 6$$\hat { j }$$ – 3$$\hat { k }$$
$$\overline { F_2 }$$ = 4$$\hat { i }$$ – 10$$\hat { j }$$ + 12$$\hat { k }$$
$$\overline { F_3 }$$ = 4$$\hat { i }$$ + 7$$\hat { j }$$
$$\overline { F }$$ = $$\overline { F_1 }$$ + $$\overline { F_2 }$$ + $$\overline { F_3 }$$
= 5$$\hat { i }$$ + 3$$\hat { j }$$ + 9$$\hat { k }$$
$$\overline { OB }$$ = 8$$\hat { i }$$ – 6$$\hat { j }$$ – 4$$\hat { k }$$
$$\overline { OA }$$ = 18$$\hat { i }$$ + 3$$\hat { j }$$ – 9$$\hat { k }$$
$$\overline { AB }$$ = $$\overline { OB }$$ – $$\overline { OA }$$
= -10$$\hat { i }$$ – 9$$\hat { j }$$ + 5$$\hat { k }$$
$$\overline { t }$$ = $$\overline { r }$$ × $$\overline { F }$$ = $$\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -10 & -9 & 5 \\ 5 & 3 & 9 \end{array}\right|$$
= $$\hat { i }$$(-81 – 15) – $$\hat { j }$$(-90 – 25) + $$\hat { k }$$(-30 + 45)
= -96$$\hat { i }$$ + 115$$\hat { j }$$ + 15$$\hat { k }$$