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:
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)
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.
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 \(\overline{\mathrm{AC}}=\overline{\mathrm{DR}}=\frac{\overline{\mathrm{c}}+\overline{\mathrm{a}}}{2}\)
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
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 \(\overline{\mathrm{AB}}, \overline{\mathrm{AE}}, \overline{\mathrm{BC}}, \overline{\mathrm{DC}}, \overline{\mathrm{ED}}\) and \(\overline{\mathrm{A C}}\) is λ \(\overline{\mathrm{A C}}\) , then find the value of λ. (S.A)
Answer:
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 \(\overline{\mathrm{AB}}\) = λ\(\overline{\mathrm{A C}}\) then find the value of λ. (March 2011) (S.A)
Answer:
Let O be the origin and given
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
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. \(\overline{\mathrm{OA}}\) = 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
∴ 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
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
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.
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:
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
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 \(\overline{\mathrm{OP}}\)
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:
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
⇒ 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:
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