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

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

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) AL and AM in terms of AB and AD
ii) λ, if AM = λ AC−AL(V.S.A)

Answer: ABCD is a parallelogram and hence AB=DC and BC=AD We have BC = 12BC (∵ L is the mid point of BC) = 1/2 AD (∵ BC = AD)

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 DA+DB+DC in terms of DP,DQ and DR.
ii) If PA+QB+RC=α, then find a.(V.S.A)

Answer: Let D be the origin. and DA¯=a , DB=b and DC=c P.V. of P is mid point of AB=DP=a+b/2 P.V. of Q is mid point of BC=DQ¯=b+c/2 P.V. of R is mid point of AC=DR¯=c+a/2

Question 3.
Let a̅ = i̅ + 2 j̅ + 3k̅ and b̅ = 3 i̅ + j̅. Find the unit vector in the direction of a+b. (V.S.A)

Answer: Unit vector in the direction of a+b is

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

Question 5.
ABCDE is a pentagon. If the sum of the vectors AB,AE,BC,DC,ED and AC is λ AC , then find the value of λ. (S.A)

Answer: Given ABCDE is a pentagon and

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 AB = λAC then find the value of λ. (March 2011) (S.A)

Answer: Let O be the origin and given

Question 7.
If OA¯=i¯+j¯+k; AB=3i¯−2j¯+k, BC=i¯+2j¯−2k and CD=2i¯+j¯+3k then find the vector OD¯. (March 2013) (V.S.A)

Answer: Since OA¯+AB+BC+CD=OD¯ ⇒ 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 ⇒ 2/4=5/m=1/n ⇒ 1/2=5/m and 1/2=1/n ⇒ m = 10 and n = 2

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:

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 ∴ 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

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. OA = i̅ – 3 j̅ + 2k̅, OB¯ = 3 i̅ – 5 j̅ + k̅ AB¯=OB¯−OA = (3i̅ – 5j̅ + k̅) – (i̅ – 3j̅ + 2k̅) = 2i̅ – 2j̅ – k̅ Let α be the angle between AB¯ and i̅ then

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

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
AB¯=OB¯−OA = (3a̅ + 2b̅ – 5c̅) – (-a̅ + 4b̅ – 3c̅)
= 4a̅ – 2b̅ – 2c̅
AB¯=OB¯−OA = (-3a̅ + 2b̅ – 5c̅) – (3a̅ + 2b̅ – 5c̅)
= -6a̅ – 4b̅ + 3c̅
AC¯=OC−OA = (-3a̅ + 8b̅ – 5c̅) – (-a̅ + 4b̅ – 3c̅) = -2a̅ + 4b̅ – 2c̅
AD=OD−OA = (3a̅ + 2b̅ + c̅) – (-a̅ + 4b̅ – 3c̅) = -2a̅ – 2b̅ + 4c̅
Let a vector be expressed as a linear combination of other two.
Suppose AB¯ = x(AC¯) + y (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
AB¯,AC¯,AD are coplanar.
and AB¯,AC¯,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.
OA = 6a̅ + 2b̅ – c̅, OB¯ = 2a̅ – b̅ + 3c̅
OC = -a̅ + 2b̅ – 4c̅, OD = -12a̅ – b̅ – 3c̅
∴ AB¯=OB¯−OA
= (2a̅ – b̅ + 3c̅) – (6a̅ + 2b̅ – c̅)
= – 4a̅ – 3b̅ + 4c̅
AC¯=OC−OA = (-a̅ + 2b̅ – 4c̅) – (6a̅ + 2b̅ – c̅) = -7a̅ – 3c̅
AD=OD−OA = (-12a̅ – b̅ – 3c̅) – (6a̅ + 2b̅ – c̅)= -18a̅ – 3b̅ – 2c̅
∴ Let a vector be expressed as a linear combination of other two.
Suppose AB¯ = xAC¯ + yAD
⇒ -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.