TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a)

Students must practice these TS Intermediate Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1A Addition of Vectors Solutions Exercise 4(a)

I.
Question 1.
ABCD is a parallelogram. If L and M are the middle points of BC CD respectively then find
i) \(\overline{\mathrm{A L}}\) and \(\overline{\mathrm{A M}}\) in terms of \(\overline{\mathrm{A B}}\) and \(\overline{\mathrm{A D}}\)
ii) λ, if \(\overline{\mathrm{A M}}\) = λ \(\overline{\mathrm{A C}}-\overline{\mathrm{A L}}\) (V.S.A)
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 1
ABCD is a parallelogram and hence \(\overline{\mathrm{AB}}=\overline{\mathrm{DC}}\) and \(\overline{\mathrm{BC}}=\overline{\mathrm{AD}}\)
We have \(\overline{\mathrm{BC}}\) = \(\frac{1}{2} \overline{\mathrm{BC}}\)
(∵ L is the mid point of BC)
= \(\frac{1}{2} \overline{\mathrm{AD}}\) (∵ BC = AD)
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 2

TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a)

Question 2.
In AABC, P, Q and R are the mid points of the sides AB, BC and CA respectively. If D is any point
i) Then express \(\overline{\mathrm{DA}}+\overline{\mathrm{DB}}+\overline{\mathrm{DC}}\) in terms of \(\overline{\mathrm{DP}}, \overline{\mathrm{DQ}}\) and \(\overline{\mathrm{DR}}\).
ii) If \(\overline{\mathrm{P A}}+\overline{\mathrm{Q B}}+\overline{\mathrm{R C}}=\bar{\alpha}\), then find a.(V.S.A)
Answer:
Let D be the origin.
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 3
and \(\overline{\mathrm{DA}}=\overline{\mathrm{a}}\) , \(\overline{\mathrm{DB}}=\overline{\mathrm{b}}\) and \(\overline{\mathrm{DC}}=\overline{\mathrm{c}}\)
P.V. of P is mid point of \(\overline{\mathrm{AB}}=\overline{\mathrm{DP}}=\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}}{2}\)
P.V. of Q is mid point of \(\overline{\mathrm{BC}}=\overline{\mathrm{DQ}}=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}\)
P.V. of R is mid point of \(\overline{\mathrm{AC}}=\overline{\mathrm{DR}}=\frac{\overline{\mathrm{c}}+\overline{\mathrm{a}}}{2}\)
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 4

Question 3.
Let a̅ = i̅ + 2 j̅ + 3k̅ and b̅ = 3 i̅ + j̅. Find the unit vector in the direction of \(\bar{a}+\bar{b}\). (V.S.A)
Answer:
Unit vector in the direction of \(\bar{a}+\bar{b}\) is
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 5

Question 4.
If the vectors -3i̅ + 4j̅ + λk̅ and μi̅ + 8j̅ + 6k̅ are collinear vectors then find λ and μ. (May 2014, ’12, Mar. ’14)
Answer:
If the vectors a1 i̅ + b1 j̅ + c1k̅ and a2 i̅ + b2 j̅ + c2k̅ are collinear then
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 6

Question 5.
ABCDE is a pentagon. If the sum of the vectors \(\overline{\mathrm{AB}}, \overline{\mathrm{AE}}, \overline{\mathrm{BC}}, \overline{\mathrm{DC}}, \overline{\mathrm{ED}}\) and \(\overline{\mathrm{A C}}\) is λ \(\overline{\mathrm{A C}}\) , then find the value of λ. (S.A)
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 7
Given ABCDE is a pentagon and
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 8

Question 6.
If the position vectors of the points A, B and C are -2i̅ + j̅ – k̅, -4i̅ + 2j̅ + 2k̅ and 6i̅ – 3j̅ – 13k̅ respectively and \(\overline{\mathrm{AB}}\) = λ\(\overline{\mathrm{A C}}\) then find the value of λ. (March 2011) (S.A)
Answer:
Let O be the origin and given
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 9

Question 7.
If \(\overline{\mathrm{OA}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}+\overline{\mathrm{k}}\); \(\overline{\mathrm{AB}}=3 \overline{\mathrm{i}}-2 \overline{\mathrm{j}}+\overline{\mathrm{k}}\), \(\overline{\mathrm{BC}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}-2 \overline{\mathrm{k}}\) and \(\overline{\mathrm{CD}}=2 \overline{\mathrm{i}}+\overline{\mathrm{j}}+3 \overline{\mathrm{k}}\) then find the vector \(\overline{\mathrm{OD}}\). (March 2013) (V.S.A)
Answer:
Since \(\overline{\mathrm{OA}}+\overline{\mathrm{AB}}+\overline{\mathrm{BC}}+\overline{\mathrm{CD}}=\overline{\mathrm{OD}}\)
⇒ \(\overline{\mathrm{OD}}\) = (i̅ + j̅ + k̅) + (3i̅ – 2j̅ + k̅) + (i̅ + 2 j̅ – 2k̅) + (2 i̅ + j̅ + 3k̅)
= 7i̅ + 2j̅ + 3k̅

Question 8.
If a̅ = 2i̅ + 5j̅ + k̅ and b̅ = 4i̅ + mj̅ + nk̅ are collinear vectors then find m and n. (May 2011) (V.S.A)
Answer:
Since a̅ = 2i̅ + 5j̅ + k̅ and
b̅ = 4i̅ + mj̅ + nk̅ are collinear
⇒ \(\frac{2}{4}=\frac{5}{m}=\frac{1}{n}\)
⇒ \(\frac{1}{2}=\frac{5}{m}\) and \(\frac{1}{2}=\frac{1}{n}\) ⇒ m = 10 and n = 2

TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a)

Question 9.
Let a̅ = 2i̅ + 4j̅ – 5k̅, b̅ = i̅ + j̅ + k̅ and c̅ = i̅ + 2k̅. Find the unit vector in the opposite direction of a̅ + b̅ + c̅. (March 2015-A.P)(May 2012; Mar. ’04, ’12; Board Model Paper) (V.S.A)
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 10

Question 10.
Is the triangle formed by the vectors 3i̅ + 5j̅ + 2k̅, 2i̅ – 3j̅ – 5k̅ and -5i̅ – 2 j̅ + 3k̅ equilateral ? (V.S.A)
Answer:
Let ABC be the triangle with AB = 3i̅ + 5 j̅ + 2k̅
BC = 2i̅ – 3 j̅ – 5k̅
CA = -5i̅ – 2j̅ + 3k̅
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 11
∴ The given vectors formed on equilateral triangle.

Question 11.
If α, β and γ are the angles made by the vector 3i̅ – 6j̅ + 2k̅ with the positive directions of the coordinate axes then find cos α, cos β, cos γ. (S.A)
Answer:
Unit vectors along the coordinate axes are respectively i̅, j̅, k̅
Let p̅ = 3i̅ – 6j̅ + 2k̅
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 12

Question 12.
Find the angles made by the straight line passing through the points (1, -3, 2) and (3, -5, 1) with the coordinate axes. (S.A)
Answer:
Let the vectors along the coordinate axes be i̅, j̅, k̅ respectively. Let O be the origin and the points A(1, -3, 2) and B(3, -5, 1).
i. e. \(\overline{\mathrm{OA}}\) = i̅ – 3 j̅ + 2k̅, \(\overline{\mathrm{OB}}\) = 3 i̅ – 5 j̅ + k̅
\(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = (3i̅ – 5j̅ + k̅) – (i̅ – 3j̅ + 2k̅) = 2i̅ – 2j̅ – k̅
Let α be the angle between \(\overline{\mathrm{AB}}\) and i̅ then
cos α = \(\frac{\overline{\mathrm{AB}} \cdot \overline{\mathrm{i}}}{|\overline{\mathrm{AB}}||\overline{\mathrm{i}}|}=\frac{2}{\sqrt{4+4+1} \cdot(1)}=\frac{2}{3}\)

Similarly if β, γ be the angles between \(\overline{\mathrm{AB}}\) and j and \(\overline{\mathrm{AB}}\) and k̅ then
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 13
∴ The angles made by the straight line AB with X, Y, Z axes are
α = cos-1\(\left(\frac{2}{3}\right)\)
β = cos-1\(\left(\frac{2}{3}\right)\)
γ = cos-1\(\left(\frac{2}{3}\right)\)

II.
Question 1.
If a̅ + b̅ + c̅ = αd̅, b̅ + c̅ + d̅ = βa̅ and a̅, b̅, c̅ are non-coplanar vectors, then show that a̅ + b̅ + c̅ + d̅ = 0̅. (S.A)
Answer:
Given a̅ + b̅ + c̅ = αd̅ ……………. (1)
b̅ + c̅ + d̅ = βa̅ …………….. (2)
From (2), d̅ = pa̅ – b̅ – c̅
From (1), a̅ + b̅ + c̅ = a, (pa̅ – b̅ – c̅)
⇒ (1 – αβ)a̅ + (1 + a)b̅ + (1 + a)c̅ = 0
∴ a̅, b̅, c̅ are non coplanar vectors
1 – αβ = 0 ⇒ αβ = 1 and
1 + α = 0 ⇒ α = -l β = -1
Hence from (1); a̅ + b̅ + c̅ = -d̅
⇒ a̅ + b̅ + c̅ + d̅ = 0

TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a)

Question 2.
a̅, b̅, c̅ are non coplanar vectors. Prove that the following four points are coplanar.
i) -a̅ + 4b̅ – 3c̅, 3a̅ + 2b̅ – 5c̅ (May,’14,’12)
-3a̅ + 8b̅ – 5c̅, – 3a̅ + 2b̅ + c̅
Answer:
Let 0 be the origin and A, B, C, D are the four points given by
OA = -a̅ + 4b̅ – 3c̅, OB = 3a̅ + 2b̅ – 5c̅
OC = -3a̅ + 8b̅ – 5c̅, OD = -3a̅ + 2b̅ + c̅
\(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = (3a̅ + 2b̅ – 5c̅) – (-a̅ + 4b̅ – 3c̅)
= 4a̅ – 2b̅ – 2c̅
\(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = (-3a̅ + 2b̅ – 5c̅) – (3a̅ + 2b̅ – 5c̅)
= -6a̅ – 4b̅ + 3c̅
\(\overline{\mathrm{AC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OA}}\) = (-3a̅ + 8b̅ – 5c̅) – (-a̅ + 4b̅ – 3c̅) = -2a̅ + 4b̅ – 2c̅
\(\overline{\mathrm{AD}}=\overline{\mathrm{OD}}-\overline{\mathrm{OA}}\) = (3a̅ + 2b̅ + c̅) – (-a̅ + 4b̅ – 3c̅) = -2a̅ – 2b̅ + 4c̅
Let a vector be expressed as a linear combination of other two.
Suppose \(\overline{\mathrm{AB}}\) = x(\(\overline{\mathrm{AC}}\)) + y (\(\overline{\mathrm{AD}}\)) where x, y are scalars.
∴ 4a̅ – 2b̅ – 2c̅ = x (-2a̅ + 4b̅ – 2c̅) + y(-2a̅ – 2b̅ + 4c̅)

Comparing coefficients of a̅, b̅, c̅ we get
(∵ a̅, b̅, c̅ are non coplanar vectors)
-2x – 2y = 4 ……………(1)
4x – 2y = -2 ……………(2)
-2x + 4y = -2 ………….(3)
Solving (1) and (2) we get 2x + 2y = – 4 and 4x – 2y = – 2
6x = – 6 ⇒ x = -1
x + y = -2 ⇒ y = -1
x = – 1 and y = -1 satisfy equation (3).
⇒ A, B, C, D are coplanar and
\(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}, \overline{\mathrm{AD}}\) are coplanar.
and \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}, \overline{\mathrm{AD}}\) are coplanar.
∴ The given points A, B, C, D are coplanar.

ii) 6a̅ + 2b̅ – c̅, 2a̅ – b̅ + 3c̅, -a̅ + 2b̅ – 4c̅, -12a̅ – b̅ – 3c̅
Answer:
Let O be the origin and A, B, C, D be the given points.
\(\overline{\mathrm{OA}}\) = 6a̅ + 2b̅ – c̅, \(\overline{\mathrm{OB}}\) = 2a̅ – b̅ + 3c̅
\(\overline{\mathrm{OC}}\) = -a̅ + 2b̅ – 4c̅, \(\overline{\mathrm{OD}}\) = -12a̅ – b̅ – 3c̅
∴ \(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\)
= (2a̅ – b̅ + 3c̅) – (6a̅ + 2b̅ – c̅)
= – 4a̅ – 3b̅ + 4c̅
\(\overline{\mathrm{AC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OA}}\) = (-a̅ + 2b̅ – 4c̅) – (6a̅ + 2b̅ – c̅) = -7a̅ – 3c̅
\(\overline{\mathrm{AD}}=\overline{\mathrm{OD}}-\overline{\mathrm{OA}}\) = (-12a̅ – b̅ – 3c̅) – (6a̅ + 2b̅ – c̅)= -18a̅ – 3b̅ – 2c̅
∴ Let a vector be expressed as a linear combination of other two.
Suppose \(\overline{\mathrm{AB}}\) = x\(\overline{\mathrm{AC}}\) + y\(\overline{\mathrm{AD}}\)
⇒ -4a̅ – 3b̅ + 4c̅ = x(-7a̅ – 3c̅) + y(-18a̅ – 3b̅ – 2c̅)

Comparing coefficients of a̅, b̅, c̅ since a̅, b̅, c̅ are non coplanar,
-7x – 18y = – 4 …………(1)
-3y = -3 ⇒ y = 1 ……….(2)
∴ -7x – 18 = – 4 ⇒ – 7x = 14 ⇒ x = -2
Comparing coefficient of c,
-3x – 2y = 4 ………..(3)
x = – 2 and y = 1 satisfy equation (3)
and hence A, B, C, D are coplanar.

Alternate Method For Above Problem :
Use scalar triple product of vectors \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}\) and \(\overline{\mathrm{AD}}\) show that this
\(\overline{\mathrm{AB}} \cdot(\overline{\mathrm{AC}} \times \overline{\mathrm{AD}})\) = 0
\([\overline{\mathrm{AB}} \overline{\mathrm{AC}} \overline{\mathrm{AD}}]=\left|\begin{array}{rrr}
-4 & -3 & 4 \\
-7 & 0 & -3 \\
-18 & -3 & -2
\end{array}\right|\)
= -4 (-9) + 3 (14 – 54) + 4 (21)
= 36- 120 + 84 = 0
∴ Vectors AB, AC, AD are coplanar
⇒ The given points A, B, C, D are coplanar.

Question 3.
If i̅, j̅, k̅ are unit vectors along the positive directions of the co-ordinate axes, then show that the four points 4i̅ + 5j̅ + k̅, – j̅ – k̅ , 3i̅ + 9j̅ + 4k̅ and -4i̅ +4j̅ +4k̅ are coplanar. (Mar. ’14)
Answer:
Let O be the origin and let A, B, C, D be the given points. Then \(\overline{\mathrm{OA}}\) = 4i̅ + 5j̅ + k̅,
\(\overline{\mathrm{OB}}\) = – j̅ – k̅, \(\overline{\mathrm{OC}}\) = 3i̅ + 9 j̅ + 4k̅,
\(\overline{\mathrm{OD}}\) = -4 i̅, + 4 j̅, + 4k̅,
Now AB = OB — OA = (-j̅ – k̅) – (4i̅ + 5j̅ + k̅) – 4i̅ – 6j̅ – 2k̅
AC = OC – OD = -i̅ + 4j̅ + 3k̅,
AD = OD – OA = -8i̅ – j̅ + 3k̅
Let \(\overline{\mathrm{AB}}\) = x(\(\overline{\mathrm{AC}}\)) + y(\(\overline{\mathrm{AD}}\)) for some values of x and y
⇒ – 4i̅ – 6j̅ – 2k̅ = x(- i̅ + 4j̅ + 3k̅) + y(-8i̅ – j̅ + 3k̅)
⇒ (x + 8y – 4) i̅ + (-4x + y – 6)j̅ + (-3x – 3y – 2)k̅ = 0
∴ i̅, j̅, k̅ are non coplanar
x + 8y – 4 = 0 …………..(1)
4x – y + 6 = 0 …………..(2)
3x + 3y + 2 = 0 ………….(3)
Solving (1) and (2) we get
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 14
Hence the vectors AB, AC and AD are coplanar
⇒ The given points A, B, C, D are coplanar.

Second method :
\(\left[\begin{array}{lll}
\overline{\mathrm{AB}} & \overline{\mathrm{AC}} & \overline{\mathrm{AD}}
\end{array}\right]=\left|\begin{array}{rrr}
-4 & -6 & -2 \\
-1 & 4 & 3 \\
-8 & -1 & 3
\end{array}\right|\)
= – 4 (12 + 3) + 6 (- 3 + 24) – 2 (1 + 32)
= – 60 + 126 – 66 = 0
The vectors \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}, \overline{\mathrm{AD}}\) are coplanar.
⇒ The given points A, B, C, D are coplanar.

TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a)

Question 4.
If a̅, b̅, c̅ are non coplanar vectors, then test for the collinearity of the following points whose position vectors are given by
(i) a̅ – 2b̅ + 3c̅, 2a̅ + 3b̅ – 4c̅, – 7b̅ + 10c̅ (S.A)
Answer:
Given a, b, c are the non coplanar vectors
Let \(\overline{\mathrm{OA}}\) = a̅ – 2b̅ + 3c̅. \(\overline{\mathrm{OB}}\) = 2a̅ + 3b̅ – 4c̅
and \(\overline{\mathrm{OC}}\) = -7b̅ + 10c̅ be the points with respect to specific origin O’.
Then \(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = a̅ + 5b̅ – 7c̅
\(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = -2a̅ – 10b̅ + 14c̅
and \(\overline{\mathrm{AC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OA}}\) = -a̅ – 5b̅ + 7c̅
∴ \(\overline{\mathrm{BC}}=-2(\overline{\mathrm{AB}}) \Rightarrow \overline{\mathrm{BC}}=2 \overline{\mathrm{BA}}\)
∴ The points A, B, C are collinear.
(∵ \(\overline{\mathrm{BC}}=\lambda \overline{\mathrm{BA}}\) where λ = 2)

ii) 3a̅ – 4b̅ + 3c̅, – 4a̅ + 5b̅ – 6c̅, 4a̅ – 7b̅ + 6c̅
Answer:
Let \(\overline{\mathrm{OA}}\) = 3a̅ – 4b̅ + 3c̅,
\(\overline{\mathrm{OB}}\) = -4a̅ + 5b̅ – 6c̅,
\(\overline{\mathrm{OC}}\) = 4a̅ – 7b̅ + 6c̅
∴ \(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = -7a̅ + 9b̅ – 9c̅
\(\overline{\mathrm{AC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OA}}\) = a̅ – 3b̅ + 3c̅
\(\overline{\mathrm{AC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OA}}\) = 8a̅ – 12b̅ + 12c̅
∴ \(\overline{\mathrm{AB}} \neq \lambda \overline{\mathrm{BC}}\), λ is a scalar.
⇒ The points A, B, C are non collinear.

iii) 2a̅ + 5b̅ – 4c̅, a̅ + 4b̅ – 3c̅, 4a̅ + 7b̅ – 6c̅
Answer:
Let O be the origin and A, B, C be the given points.
Then \(\overline{\mathrm{OA}}\) = 2a̅ + 5b̅ – 4c̅.
\(\overline{\mathrm{OB}}\) = a̅ + 4b̅ – 3c̅.
and \(\overline{\mathrm{OC}}\) = 4a̅ + 7b̅ – 6c̅
\(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = (a̅ + 4b̅ – 3c̅) – (2a̅ + 5b̅ – 4c̅)
= -a̅ – b̅ + c̅
\(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = (4a̅ + 7b̅ – 6c̅) – (a̅ + 4b̅ – 3c̅)
= 3a̅ + 3b̅ – 3c̅ = -3(a̅ – b̅ + c̅)
\(\overline{\mathrm{BC}}=-3(\overline{\mathrm{AB}})\)
⇒ \(\overline{\mathrm{BC}}=3(\overline{\mathrm{BA}})\) where λ = 3
∴ The points A, B, C are collinear.

III.
Question 1.
In the cartesian plane, O is the origin of the coordinate axes. A person starts at O and walks a distance of 3 units in the North-East direction and reaches the point P. From P he walks 4 units of distance parallel to North-West direction and reaches the point
Q. Express the vector \(\overline{\mathrm{OQ}}\) in terms of i̅ and j̅ (observe that (∠XOP=45°) (S.A)
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 15
O is the origin and ∠XOP = 45°
The person starts at 0 and walks a distance of 3 units in North-East direction.
∴ \(\overline{\mathrm{OP}}\) = (3cos45°) i̅ + (3sin45°) j̅
= \(\frac{3}{\sqrt{2}}\)i̅ + \(\frac{3}{\sqrt{2}}\)j̅
PQ = 4 units
and pp is parallel to X axis
∴ ∠RPQ = 135°
PQ is parallel to North-West direction
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 16

Question 2.
The points O, A, B, X and Y are such that \(\overline{\mathrm{OA}}\) = a̅, \(\overline{\mathrm{OB}}\) = b̅, \(\overline{\mathrm{OX}}\) = 3a̅ and \(\overline{\mathrm{OY}}\) = 3b̅. Find \(\overline{\mathrm{BX}}\) and \(\overline{\mathrm{AY}}\) interms of a and 5. Further, if the point P divides AY in the ratio 1 : 3, then express \(\overline{\mathrm{BP}}\) in terms of a and b. (S.A)
Answer:
Given \(\overline{\mathrm{OA}}\) = a̅, \(\overline{\mathrm{OB}}\) = b̅, \(\overline{\mathrm{OX}}\) = 3a̅ \(\overline{\mathrm{OY}}\) = 3b̅
\(\overline{\mathrm{BX}}=\overline{\mathrm{OX}}-\overline{\mathrm{OB}}\) = 3a̅ – b̅
\(\overline{\mathrm{AY}}=\overline{\mathrm{OY}}-\overline{\mathrm{OA}}\) = 3b̅ – a̅
If P divides \(\overline{\mathrm{AY}}\) in the ratio 1 : 3 then the position vector of P is \(\overline{\mathrm{OP}}\)
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 17

TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a)

Question 3.
In ΔOAB, E is the midpoint of AB and F Is a point on OA such that OF = 2 FA. If C Is the point of intersection of \(\overline{\mathrm{OE}}\) and \(\overline{\mathrm{BF}}\) then find the ratios OC : CE and BC : CF. (S.A)
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 18
Let O be the origin and \(\overline{\mathrm{OA}}\) = a̅, \(\overline{\mathrm{OB}}\) = b̅
Since E is the midpoint of AB,
\(\overline{\mathrm{OE}}\) = \(\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}}{2}\)

and OF = 2 FA ⇒ F divides OA in the ratio 2 : 1
\(\overline{\mathrm{OF}}\) = \(\frac{2 \bar{a}+1(0)}{2+1}=\frac{2}{3} \bar{a}\)

C is the point of intersection of \(\overline{\mathrm{OE}}\) and \(\overline{\mathrm{BF}}\).
let C divides \(\overline{\mathrm{OE}}\) in the ratio 1 : λ then the position vector of C is
\(\overline{\mathrm{OC}}\) = \(\frac{1(\overline{\mathrm{OE}})+\lambda(0)}{1+\lambda}=\frac{\overline{\mathrm{OE}}}{\lambda+1}=\frac{\bar{a}+\bar{b}}{2(\lambda+1)}\) ………..(1)

Let C divides \(\overline{\mathrm{BF}}\) in the ratio μ : 1 then the position vector of C is
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 19
⇒ 4(λ + 1) = 5
⇒ 4λ = 1 ⇒ λ = \(\frac{1}{4}\)

C divides OE in the ratio 1 : \(\frac{1}{4}\) = 4 : 1
∴ OC : CE = 4 : 1
C divides BF in the ratio μ : 1 = \(\frac{3}{2}\) : 1
= 3 : 2
∴ BC : CF = 3 : 2

Question 4.
The point E divides the segment PQ internally in the ratio 1 : 2 and R is any point not on the line PQ. If F is a point on QR such that QF: FR = 2 : 1, then show that EF is parallel to PR. (S.A)
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 20
Let O be the origin and \(\overline{\mathrm{OP}}=\overline{\mathrm{a}}, \overline{\mathrm{OQ}}=\overline{\mathrm{b}}\) and \(\overline{\mathrm{OP}}=\overline{\mathrm{a}}, \overline{\mathrm{OQ}}=\overline{\mathrm{b}}\)
E divides PQ in the ratio 1: 2
TS Inter 1st Year Maths 1A Solutions Chapter 4 Addition of Vectors Ex 4(a) 21

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