ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.5

Students can cross-reference their work with ISC Mathematics Class 12 Solutions to ensure accuracy.

ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.5

Typical Problems:

Question 1.
If M and N are the mid-points of the sides OA and BC of a parallelogram OABC, prove that CM and AN cut the diagonal OB in its points of trisection which are also the points of trisection of CM and AN respectively.
Solutions:
Taking O as origin, let \(\vec{a} \text { and } \vec{c}\) be the position vector of A and C.
∴ P.V of B = \(\overrightarrow{\mathrm{OB}}\)
= \(\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}}\)
= \(\vec{a}+\vec{c}\)
Since M and N are the mid points ofsîdes OA and BC of ||gm OABC.

∴ P.V of M = \(\frac{\vec{a}}{2}\)
and P.V of N = \(\overrightarrow{\mathrm{ON}}\)
= \(\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{CN}}\)
= \(\vec{c}+\frac{\vec{a}}{2}\)
Thus the position vector of a point dividing OB in the 1 : 2 = \(\frac{1(\vec{a} \cdot \vec{c})+2 \cdot \overrightarrow{0}}{1+2}\)
= \(\frac{\vec{a}+\vec{c}}{3}\)
Also, the position vector of point dividing CM in the ratio 2 : 1 = \(\frac{2\left(\frac{\vec{a}}{2}\right)+1(\vec{c})}{2+1}\)
= \(\frac{\vec{a}+\vec{c}}{3}\)
Hence, the point of trisection of OB is ame as the poInt of trisection of CM.
Hence CM and OB in its points of trisection which is also the point of section of CM.
SiinilarlyAN cuts the diagonal OB in its point oftrisection which is also the point of trisection of AN.

Question 2.
ABCD ¡s a parallelogram and E is the mid-point of CD. If F is a point on AE such that it divides AE internally in ratio 2 : 1, show by vector method that F lies on the diagonal BD and that it divides BD internally in ratio 2 : 1.
Solutions:
Taking A as origin, let \(\vec{b} \text { and } \vec{d}\) are the position vectors of vertices B and D of ||gm ABCD.
∴ P.V of C = \(\overrightarrow{\mathrm{AC}}\)
= \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}\)
= \(\vec{b}+\vec{d}\)

Since E be the mid point of CD.
∴ P.V of E = \(\frac{\vec{d}+\vec{b}+\vec{d}}{2}\)
= \(\frac{2 \vec{d}+\vec{b}}{2}\)
Since F be a point on AE such that F divides AE internally in the ratio 2 : 1
∴ P.V of F = \(\frac{2\left(\frac{2 \vec{d}+\vec{b}}{2}\right)+1 \times \overrightarrow{0}}{2+1}\)
= \(\frac{2 \vec{d}+\vec{b}}{3}\)
Also P.V of a point say G that divides diagonal BD internally in the ratio 2 : 1
= \(\frac{1(\vec{b})+2 \vec{d}}{1+2}\)
= \(\frac{\vec{b}+2 \vec{d}}{3}\)
Thus the point G coincides with point F.
Hence F lies on diagonal BD and dividing BD internally in the ratio 2 : 1.

Question 3.
In a triangle OAB, E is mid-point of OB and D is a point on AB such that AD: DB = 2 : 1. If OD and AE intersect at P, determine the ratio OP : Pi), using vector methods.
Solution:
Taking A as origin and \(\vec{a} \text { and } \vec{b}\) are the position vector of vertices B andC of zOAB.
∴ \(\overrightarrow{\mathrm{OA}}=\vec{a}\) ;
\(\overrightarrow{\mathrm{OB}}=\vec{b}\)
Since E be the mid point of OB.

∴ P.V of E = \(\frac{\vec{b}}{2}\)
∴ Since AD : DB = 2 : 1
∴ D divides AB in the ratio 2 : 1
∴ P.V of D = \(\frac{2 \vec{b}+\vec{a}}{3}\).
∴Since P lies on AE
∴ A, P and E are collinear.
∴ \(\overrightarrow{\mathrm{AP}}=t \overrightarrow{\mathrm{AE}}\) for some non-zero scalar t

⇒ 2λ = 3
⇒ λ = \(\frac{3}{2}\)
Thus P divides OD in the ratio 3 : 2 internally.
∴ OP : PD = 3 : 2

Question 4.
ABCD is a parallelogram. P and Q are points on the sides AB and AD respectively such that AB 2AP and AD = 3AQ. Show that the diagonal AC divides PQ ¡n the ratio 3 : 2 and is itself divided by PQ in the ratio 1 : 4.
Solution:
Taking A as origin and \(\vec{b} \text { and } \vec{d}\) are the position vectors of points B and D of gm ABCD.
∴ P.V 0f C = \(\overrightarrow{\mathrm{AC}}\)
= \(\vec{b}+\vec{d}\)

Since P be a point on AB s.tSince P be a point on AB s.l AB = 2AP
⇒ AP + PB = 2AP
⇒ PB = AP
∴ P.V of P = \(\frac{\vec{b}}{2}\)
Also Q be a point on AD s.t AD = 3AQ
⇒ AQ + QD = 3AQ
⇒ QD = 2AQ
= \(\frac{\mathrm{QD}}{\mathrm{AQ}}=\frac{2}{1}\)
Thus Q divides DA in the ratio 2 : 1.
∴ P.V of Q = \(\frac{\vec{d}}{3}\)
Let F divides AC in the ratio λ : 1
and QP in the ratio μ : 1.
∴ \(\frac{\lambda(\vec{b}+\vec{d})}{\lambda+1}=\frac{\frac{\mu \vec{b}}{2}+\frac{\vec{d}}{3}}{\mu+1}\)

Therefore, F divides AC in the ratio 1 : 4 and
F divides PQ in the ratio 3 : 2.
Thus, diagonal AC divides PQ in the ratio 3 : 2 and AC itself divided by PQ in the ratio 1 : 4.

Question 5.
Prove by vector method that the line segment joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and equal to half of their difference.
Solution:
Let ABCD be the trapezium
and let \(\vec{a}, \vec{b}, \vec{c}\) and \(\vec{d}\) are the position vectors of vertices
A, B, C and D of trapezium with reference to some origin O.
Let L and M are the mid points of diagonals BD and AC.

∴ from (2) and (3) ; we have
\(|\overrightarrow{\mathrm{LM}}|=\frac{1}{2}\{|\overrightarrow{\mathrm{DC}}|-|\overrightarrow{\mathrm{AB}}|\}\)
Hence, the line segment joining the mid-points of the diagonals of a trapezium is parallel to parallel sides and equal to half of their difference.

Question 6.
In a triangle ABC, D and E are points on BC and AC respectively, such that BD = 2DC and AE = 3EC. Let P be the point of intersection of AD and BE. Find BP / PE using vector method.
Solution:
Taking A as origin and let \(\vec{b} \text { and } \vec{c}\) are the position vectors of B and C respectively.
Since D he a point on BC such that
BD = 2DC
⇒ \(\frac{\mathrm{BD}}{\mathrm{DC}}\) = 2
∴ D divides BC in the ratio 2 : 1.
∴ P.V. of D = \(\frac{2 \vec{c}+\vec{d}}{3}\)

Since E be a point on AC s.t AE = 3EC
So E divides AC in the ratio 3 : 1.
∴ P.V of E = \(\overrightarrow{\mathrm{AE}}\)
= \(\frac{3 \vec{c}}{4}\)
Now P be the point of intersection of AD
and BE and divides BE in the ratio λ : 1
∴ P.V of P = \(\frac{\lambda\left(\frac{3 \vec{c}}{4}\right)+\vec{b}}{\lambda+1}\)
Also P lies on AD
∴ A, P and D are collinear
So \(\overrightarrow{\mathrm{AP}}\) P.V of P = t \(\overrightarrow{\mathrm{AD}}\) for some scalar t
⇒ \(\overrightarrow{\mathrm{AP}}=t\left(\frac{2 \vec{c}+\vec{b}}{3}\right)\)

Question 7.
If \(\vec{a} \cdot \vec{b}\) ≠ 0, find the vector \(\vec{r}\) which satisfies the equation : \(\vec{r} \times \vec{b}=\vec{c} \times \vec{b}\) and \(\vec{r} \cdot \vec{a}\) = 0.
Solution:

Question 8.
If \(\vec{a}, \vec{b}, \vec{c}\) are unit vectors such that \(\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}\) = 0 and the angle between \(\vec{b} \text { and } \vec{c}\) is \(\frac{\pi}{3}\), then prove taht \(\vec{a}= \pm 2(\vec{b} \times \vec{c})\).
Solution:

Question 9.
If O and H are respectively the circumcentre and orthocentre of a triangle whose vertices are the points A, B and C with position vectors \(\vec{a}, \vec{b} \text { and } \vec{c}\) respectively with reference to O as origin, then prove that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{\mathrm{OH}}\).
Solution:
Let G be the centroid ofABC.
First of all we show that circumcentre O, centroid G and orthocentre H are collinear and 2OG = HG.
Let AL and BM are altitudes drawn on sides BC and CA of ∆ABC and
let both these attitudes intersects at H.
Let AD be the median drawn on side BC.
From O, draw OD, ⊥ from O on side BC.
Let R be the circum radius of circumcircle of ∆ABC.
Thrn OB = OC = R

In ∆OBD, we have OD = R cos A ……….(1)
In ∆ABM,
we have AM = AB cos A = c cos A …………(2)
ln ∆ AHM
we have AH = AM sec (90° — C)
⇒ AH = \(\frac{c^{\prime} \cos \mathrm{A}}{\sin C}\)
[Using sine formula \(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = 2R]
⇒ AH = 2R cos A = 2OD [Using (1)]
Now ∆‘s AGH and OGD are similar
\(\frac{\mathrm{OG}}{\mathrm{HG}}=\frac{\mathrm{GD}}{\mathrm{GA}}=\frac{\mathrm{OD}}{\mathrm{AH}}=\frac{1}{2}\)
⇒ 2OG = HG ………………….(3)
Since \(\overrightarrow{\mathrm{SA}}+\overrightarrow{\mathrm{SB}}+\overrightarrow{\mathrm{SC}}=\overrightarrow{\mathrm{SA}}+2 \overrightarrow{\mathrm{SD}}\)
where S be any point in plane of ∆ ABC and D be the mid point of BC.
⇒ \(\overrightarrow{\mathrm{SA}}+\overrightarrow{\mathrm{SB}}+\overrightarrow{\mathrm{SC}}=(1+2) \overrightarrow{\mathrm{SG}}\)
Since G divides AD in the ratio 2 : 1.
\(\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}=3 \overrightarrow{\mathrm{OG}}=2 \overrightarrow{\mathrm{OG}}+\overrightarrow{\mathrm{OG}}\)
= \(\overrightarrow{\mathrm{OG}}+\overrightarrow{\mathrm{HG}}\) [using (3)]
= \(\overrightarrow{\mathrm{OG}}+\overrightarrow{\mathrm{GH}}=\overrightarrow{\mathrm{OH}}\)

Question 10.
If \(\left|\begin{array}{lll}
a & a^2 & 1+a^3 \\
b & b^2 & 1+b^3 \\
c & c^2 & 1+c^3
\end{array}\right|\) = 0 and the vectors \(\overrightarrow{\mathbf{A}}\) = (1, a, a²), \(\overrightarrow{\mathbf{B}}\) = (1, b, b²), \(\overrightarrow{\mathbf{C}}\) = (1, c, c) are non-coplanar, then prove that abc = – 1.
Solution: