NCERT Exemplar - Vectors Commerce Notes | EduRev

JEE Revision Notes

Commerce : NCERT Exemplar - Vectors Commerce Notes | EduRev

The document NCERT Exemplar - Vectors Commerce Notes | EduRev is a part of the Commerce Course JEE Revision Notes.
All you need of Commerce at this link: Commerce

SHORT ANSWER TYPE QUESTIONS

Q.1. Find the unit vector in the direction of sum of vectors
NCERT Exemplar - Vectors Commerce Notes | EduRev
Ans.
Given that
NCERT Exemplar - Vectors Commerce Notes | EduRev
∴ Unit vector in the direction ofNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required unit vector isNCERT Exemplar - Vectors Commerce Notes | EduRev

Q.2. IfNCERT Exemplar - Vectors Commerce Notes | EduRevfind the unit vector in the direction of
(i) NCERT Exemplar - Vectors Commerce Notes | EduRev
(ii)NCERT Exemplar - Vectors Commerce Notes | EduRev
Ans.
Given thatNCERT Exemplar - Vectors Commerce Notes | EduRev
(i) NCERT Exemplar - Vectors Commerce Notes | EduRev
∴ Unit vector in the direction ofNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required unit vector isNCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
(ii)NCERT Exemplar - Vectors Commerce Notes | EduRev-NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
∴ Unit vector in the direction ofNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required unit vector isNCERT Exemplar - Vectors Commerce Notes | EduRev

Q.3. Find a unit vector in the direction ofNCERT Exemplar - Vectors Commerce Notes | EduRevwhere P and Q have co-ordinates (5, 0, 8) and (3, 3, 2), respectively.
Ans.
Given coordinates are P(5, 0, 8) and Q(3, 3, 2)
NCERT Exemplar - Vectors Commerce Notes | EduRev
∴ Unit vector in the direction ofNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required unit vector isNCERT Exemplar - Vectors Commerce Notes | EduRev

Q.4. If NCERT Exemplar - Vectors Commerce Notes | EduRevare the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that BC = 1.5 BA.
Ans.
Given that
BC = 1.5 BA
NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required vector isNCERT Exemplar - Vectors Commerce Notes | EduRev

Q.5. Using vectors, find the value of k such that the points (k, – 10, 3), (1, –1, 3) and (3, 5, 3) are collinear.
Ans.
Let the given points are A( k , - 10 , 3), B(1,- 1, 3) and C(3, 5, 3)
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev

If A, B and C are collinear, then

NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Squaring both sides, we have
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
= 9 + k2- 6k + 225
NCERT Exemplar - Vectors Commerce Notes | EduRev
Dividing by 2, we get
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev(Dividing by 2)
Squaring both sides, we get
⇒ 10(k2 – 2k + 82) = 784 + k2 – 56k
⇒ 10k2 – 20k + 820 = 784 + k2 – 56k
⇒ 10k2 – k2 – 20k + 56k + 820 – 784 = 0
⇒ 9k2 + 36k + 36 = 0
⇒ k2 + 4k + 4 = 0
⇒ (k + 2)2 = 0
⇒ k + 2 = 0
⇒ k = – 2
Hence, the required value is k = – 2

Q.6. A vectorNCERT Exemplar - Vectors Commerce Notes | EduRevis inclined at equal angles to the three axes. If the magnitude of NCERT Exemplar - Vectors Commerce Notes | EduRev is 2 √3 units, findNCERT Exemplar - Vectors Commerce Notes | EduRev
Ans.
Since, the vectorNCERT Exemplar - Vectors Commerce Notes | EduRevmakes equal angles with the axes, their direction cosines should be same
∴ l = m = n
We know that
l2 + m2 + n2 = 1
⇒ l2 + l2 + l2 = 1
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
We know thatNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required value ofNCERT Exemplar - Vectors Commerce Notes | EduRev

Q.7. A vector NCERT Exemplar - Vectors Commerce Notes | EduRevhas magnitude 14 and direction ratios 2, 3, – 6. Find the direction cosines and components of NCERT Exemplar - Vectors Commerce Notes | EduRev , given that NCERT Exemplar - Vectors Commerce Notes | EduRevmakes an acute angle with x-axis.
Ans.
LetNCERT Exemplar - Vectors Commerce Notes | EduRevbe three vectors such thatNCERT Exemplar - Vectors Commerce Notes | EduRevandNCERT Exemplar - Vectors Commerce Notes | EduRev
If l, m and n are the direction cosines of vector NCERT Exemplar - Vectors Commerce Notes | EduRev, then
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
We know that l2 + m2 + n= 1
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
∴ k = ± 2 and l =NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required direction cosines areNCERT Exemplar - Vectors Commerce Notes | EduRevand the components of
NCERT Exemplar - Vectors Commerce Notes | EduRev

Q.8. Find a vector of magnitude 6, which is perpendicular to both the vectors NCERT Exemplar - Vectors Commerce Notes | EduRevandNCERT Exemplar - Vectors Commerce Notes | EduRev
Ans.
LetNCERT Exemplar - Vectors Commerce Notes | EduRevandNCERT Exemplar - Vectors Commerce Notes | EduRev
We know that unit vector perpendicular toNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Now the vector of magnitude 6 =NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required vector isNCERT Exemplar - Vectors Commerce Notes | EduRev

Q.9. Find the angle between the vectorsNCERT Exemplar - Vectors Commerce Notes | EduRev.
Ans.
LetNCERT Exemplar - Vectors Commerce Notes | EduRev and let θ be the angle betweenNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required value of θ isNCERT Exemplar - Vectors Commerce Notes | EduRev

Q.10. IfNCERT Exemplar - Vectors Commerce Notes | EduRev show that NCERT Exemplar - Vectors Commerce Notes | EduRev  Interpret the result geometrically?
Ans.
Given thatNCERT Exemplar - Vectors Commerce Notes | EduRev
So,NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev...(i)
NowNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev...(ii)
From eq. (i) and (ii) we get
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence proved.
Geometrical Interpretation
According to figure, we have
Area of parallelogram ABCD is

NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
Since, the parallelograms on the same base and between the same parallel lines are equal in area
NCERT Exemplar - Vectors Commerce Notes | EduRev

Q.11. Find the sine of the angle between the vectorsNCERT Exemplar - Vectors Commerce Notes | EduRev andNCERT Exemplar - Vectors Commerce Notes | EduRev
Ans.
Given thatNCERT Exemplar - Vectors Commerce Notes | EduRev
We know thatNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence,NCERT Exemplar - Vectors Commerce Notes | EduRev

Q.12. If A, B, C, D are the points with position vectorsNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRevrespectively, find the projection ofNCERT Exemplar - Vectors Commerce Notes | EduRev
Ans.
Here, Position vector of A =NCERT Exemplar - Vectors Commerce Notes | EduRev
Position vector of B =NCERT Exemplar - Vectors Commerce Notes | EduRev
Position vector of C =NCERT Exemplar - Vectors Commerce Notes | EduRev
Position vector of D =NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Projection ofNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required projection = √21 .

Q.13. Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, – 1, 4) and C(4, 5, – 1).
Ans.
Given that A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1)
NCERT Exemplar - Vectors Commerce Notes | EduRev
Area of ΔABC =NCERT Exemplar - Vectors Commerce Notes | EduRev=NCERT Exemplar - Vectors Commerce Notes | EduRev
=NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required area isNCERT Exemplar - Vectors Commerce Notes | EduRev

Q.14. Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.
Ans.
Let ABCD and ABFE be two parallelograms on the same base AB and between same parallel lines AB and DF.
NCERT Exemplar - Vectors Commerce Notes | EduRevLetNCERT Exemplar - Vectors Commerce Notes | EduRev
∴ Area of parallelogram ABCD =NCERT Exemplar - Vectors Commerce Notes | EduRev
Now Area of parallelogram ABFE =NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence proved.


LONG ANSWER TYPE QUESTIONS

Q.15. Prove that in any triangle ABC,NCERT Exemplar - Vectors Commerce Notes | EduRevwhere a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.
Ans.
Here, in the given figure, the components of c are c cos A and c sin A.
NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
In DBDC,
a2 = CD2 + BD2
⇒ a2 = (b – c cos A)+ (c sin A)2
⇒ a2 = b2 + c2 cos2 A – 2bc cos A + c2 sin2 A
⇒ a2 = b2 + c2(cos2 A + sin2 A) – 2bc cos A
⇒ a2 = b2 + c2 – 2bc cos A
⇒ 2bc cos A = b2 + c2 - a2
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence Proved.

Q.16. IfNCERT Exemplar - Vectors Commerce Notes | EduRevdetermine the vertices of a triangle, show that
NCERT Exemplar - Vectors Commerce Notes | EduRevgives the vector area of the triangle. Hence deduce the condition that the three pointsNCERT Exemplar - Vectors Commerce Notes | EduRev are collinear. Also find the unit vector normal to the plane of the triangle.
Ans.
Since,NCERT Exemplar - Vectors Commerce Notes | EduRevare the vertices of ΔABC
NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
For three vectors are collinear, area of ΔABC = 0
NCERT Exemplar - Vectors Commerce Notes | EduRev
which is the condition of collinearity ofNCERT Exemplar - Vectors Commerce Notes | EduRev
LetNCERT Exemplar - Vectors Commerce Notes | EduRevbe th e unit vector normal to the plane of the ΔABC
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev

Q.17. Show that area of the parallelogram whose diagonals are given by NCERT Exemplar - Vectors Commerce Notes | EduRevisNCERT Exemplar - Vectors Commerce Notes | EduRev. Also find the area of the parallelogram whose diagonals are NCERT Exemplar - Vectors Commerce Notes | EduRevandNCERT Exemplar - Vectors Commerce Notes | EduRev
Ans.
Let ABCD be a parallelogram such that,
NCERT Exemplar - Vectors Commerce Notes | EduRev
∴ by law of triangle, we get
NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
Adding eq. (i) and (ii) we get,
NCERT Exemplar - Vectors Commerce Notes | EduRev
Subtracting eq. (ii) from eq. (i) we get
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
So, the area of the parallelogram ABCD =NCERT Exemplar - Vectors Commerce Notes | EduRev
Now area of parallelogram whose diagonals areNCERT Exemplar - Vectors Commerce Notes | EduRevandNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required area isNCERT Exemplar - Vectors Commerce Notes | EduRev

Q.18. IfNCERT Exemplar - Vectors Commerce Notes | EduRevfind a vector NCERT Exemplar - Vectors Commerce Notes | EduRev such that  NCERT Exemplar - Vectors Commerce Notes | EduRev
Ans.
LetNCERT Exemplar - Vectors Commerce Notes | EduRev
Also given thatNCERT Exemplar - Vectors Commerce Notes | EduRev
Since,NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev

On comparing the like terms, we get
c3 – c2 = 0 ...(i)
c1 – c3 = 1 ...(ii)

and c2 – c1 = –1 ...(iii)
Now
NCERT Exemplar - Vectors Commerce Notes | EduRev

NCERT Exemplar - Vectors Commerce Notes | EduRev

∴ c1 + c2 + c3 = 3 ...(iv)
Adding eq. (ii) and eq. (iii) we get,
c2 – c3 = 0 ...(v)
From (iv) and (v) we get
c1 + 2c2 = 3 ...(vi)
From (iii) and (vi) we get
Adding
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
c3 – c2 = 0
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Now c2 – c1 = – 1 ⇒NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence,NCERT Exemplar - Vectors Commerce Notes | EduRev


OBJECTIVE TYPE QUESTIONS

Q.19. The vector in the direction of the vectorNCERT Exemplar - Vectors Commerce Notes | EduRevthat has magnitude 9 is
(a)NCERT Exemplar - Vectors Commerce Notes | EduRev
(b)NCERT Exemplar - Vectors Commerce Notes | EduRev
(c)NCERT Exemplar - Vectors Commerce Notes | EduRev
(d)NCERT Exemplar - Vectors Commerce Notes | EduRev
Ans. (c)
Solution.
LetNCERT Exemplar - Vectors Commerce Notes | EduRev
Unit vector in the direction ofNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
∴ Vector of magnitude 9 =NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (c).

Q.20. The position vector of the point which divides the join of points
NCERT Exemplar - Vectors Commerce Notes | EduRevin the ratio 3 : 1 is
(a)NCERT Exemplar - Vectors Commerce Notes | EduRev
(b)NCERT Exemplar - Vectors Commerce Notes | EduRev
(c)NCERT Exemplar - Vectors Commerce Notes | EduRev
(d)NCERT Exemplar - Vectors Commerce Notes | EduRev
Ans. (d)
Solution.
The given vectors areNCERT Exemplar - Vectors Commerce Notes | EduRevand the ratio is 3 : 1.
∴ The position vector of the required point c which divides the join of the given vectors NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (d).

Q.21. The vector having initial and terminal points as (2, 5, 0) and (–3, 7, 4), respectively is
(a)NCERT Exemplar - Vectors Commerce Notes | EduRev
(b)NCERT Exemplar - Vectors Commerce Notes | EduRev
(c)NCERT Exemplar - Vectors Commerce Notes | EduRev
(d)NCERT Exemplar - Vectors Commerce Notes | EduRev
Ans. (c)
Solution.
Let A and B be two points whose coordinates are given as (2, 5, 0) and (– 3, 7, 4)
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (c).

Q.22. The angle between two vectorsNCERT Exemplar - Vectors Commerce Notes | EduRevwith magnitudes √3 and 4, respectively, andNCERT Exemplar - Vectors Commerce Notes | EduRev
(a) π/6
(b) π/3
(c) π/2
(d) 5π/2
Ans. (b)
Solution.
Here, given thatNCERT Exemplar - Vectors Commerce Notes | EduRev
∴ From scalar product, we know that
NCERT Exemplar - Vectors Commerce Notes | EduRev

NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (b).

Q.23. Find the value of λ such that the vectorsNCERT Exemplar - Vectors Commerce Notes | EduRev

are orthogonal
(a) 0 
(b) 1 
(c) 3/2
(d) -5/2
Ans. (d)
Solution.
Given thatNCERT Exemplar - Vectors Commerce Notes | EduRev
SinceNCERT Exemplar - Vectors Commerce Notes | EduRevare orthogonal  
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
⇒ 2 + 2λ + 3 = 0
⇒ 5 + 2λ = 0 ⇒NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (d).

Q.24. The value of λ for which the vectorsNCERT Exemplar - Vectors Commerce Notes | EduRev
are parallel is
(a) 2/3
(b) 2/3
(c) 5/2
(d) 2/5
Ans. (a)
Solution.

Let
NCERT Exemplar - Vectors Commerce Notes | EduRev
Since the given vectors are parallel,
∴ Angle between them is 
soNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Squaring both sides, we get
900 + λ2 + 60λ = 46(20 + λ2)
⇒ 900 + λ2 + 60λ = 920 + 46λ2
⇒ λ2 – 46λ2 + 60λ + 900 – 920 = 0
⇒ - 45λ2 + 60λ - 20 = 0
⇒ 9λ2 – 12λ + 4 = 0
⇒(3λ – 2)2 = 0
⇒3λ – 2 = 0
⇒ 3λ = 2
∴ λ = 2/3
Alternate method:
LetNCERT Exemplar - Vectors Commerce Notes | EduRev
IfNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (a).

Q.25. The vectors from origin to the points A and B are
NCERT Exemplar - Vectors Commerce Notes | EduRev,respectively, then the area of triangle OAB is   
(a) 340 
(b) √25 
(c) √229 
(d)NCERT Exemplar - Vectors Commerce Notes | EduRev
Ans. (d)
Solution.

Let O be the origin
NCERT Exemplar - Vectors Commerce Notes | EduRev
andNCERT Exemplar - Vectors Commerce Notes | EduRev
∴ Area of ΔOAB =NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence the correct option is (d).

Q.26. For any vectorNCERT Exemplar - Vectors Commerce Notes | EduRevthe value ofNCERT Exemplar - Vectors Commerce Notes | EduRevis equal to
(a)NCERT Exemplar - Vectors Commerce Notes | EduRev
(b)NCERT Exemplar - Vectors Commerce Notes | EduRev
(c)NCERT Exemplar - Vectors Commerce Notes | EduRev
(d)NCERT Exemplar - Vectors Commerce Notes | EduRev
Ans. (d)
Solution.

LetNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Now,NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
SimilarlyNCERT Exemplar - Vectors Commerce Notes | EduRev
andNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (d).

Q.27. IfNCERT Exemplar - Vectors Commerce Notes | EduRevthen value ofNCERT Exemplar - Vectors Commerce Notes | EduRevis
(a) 5 

(b) 10 
(c) 14 
(d) 16
Ans. (d)
Solution.

Given thatNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev

⇒ 12 = 10 × 2 × cos θ
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
NowNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (d).

Q.28. The vectorsNCERT Exemplar - Vectors Commerce Notes | EduRevare coplanar if
(a) λ = –2 
(b) λ = 0 
(c) λ = 1 
(d) λ = – 1
Ans. (a)
Solution.

LetNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
IfNCERT Exemplar - Vectors Commerce Notes | EduRevare coplanar, then
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev

⇒ l(λ2 – 1) – 1 (λ + 2) + 2(–1 – 2l) = 0
⇒ λ3 – λ – λ – 2 – 2 – 4λ = 0
⇒ λ3 – 6λ – 4 = 0
⇒ (λ + 2) (λ2 – 2λ – 2) = 0
⇒ λ = – 2 or λ2 – 2λ – 2 = 0
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (a).

Q.29. IfNCERT Exemplar - Vectors Commerce Notes | EduRevare unit vectors such thatNCERT Exemplar - Vectors Commerce Notes | EduRevthen the value of
NCERT Exemplar - Vectors Commerce Notes | EduRev

(a) 1
(b) 3
(c) -3/2

(d) None of these
Ans. (c)
Solution.

Given thatNCERT Exemplar - Vectors Commerce Notes | EduRev
andNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (c).

Q.30. Projection vector ofNCERT Exemplar - Vectors Commerce Notes | EduRevis
(a)NCERT Exemplar - Vectors Commerce Notes | EduRev
(b)NCERT Exemplar - Vectors Commerce Notes | EduRev
(c)NCERT Exemplar - Vectors Commerce Notes | EduRev
(d)NCERT Exemplar - Vectors Commerce Notes | EduRev
Ans. (a)
Solution.

The projection vector ofNCERT Exemplar - Vectors Commerce Notes | EduRev NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (a).

Q.31. IfNCERT Exemplar - Vectors Commerce Notes | EduRevare three vectors such thatNCERT Exemplar - Vectors Commerce Notes | EduRev
then value ofNCERT Exemplar - Vectors Commerce Notes | EduRev
(a) 0 
(b) 1 
(c) – 19 
(d) 38
Ans. (c)
Solution.

Given thatNCERT Exemplar - Vectors Commerce Notes | EduRev    
andNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (c).

Q.32. IfNCERT Exemplar - Vectors Commerce Notes | EduRevand −3 ≤ λ ≤ 2 , then the range ofNCERT Exemplar - Vectors Commerce Notes | EduRevis   
(a) [0, 8]
(b) [– 12, 8]     
(c) [0, 12]     
(d) [8, 12]

Ans. (b)
Solution.

Given thatNCERT Exemplar - Vectors Commerce Notes | EduRev
NowNCERT Exemplar - Vectors Commerce Notes | EduRev
Here - 3 ≤ λ ≤ 2
⇒ - 3.4 ≤ 4λ  ≤ 2.4
⇒ - 12 ≤ 4λ  ≤ 8
∴ 4λ  = [- 12, 8]
Hence, the correct option is (b).

Q.33. The number of vectors of unit length perpendicular to the vectorsNCERT Exemplar - Vectors Commerce Notes | EduRevandNCERT Exemplar - Vectors Commerce Notes | EduRev
(a) one 
(b) two 
(c) three 
(d) infinite
Ans. (b)
Solution.

The number of vectors of unit length perpendicular to vectors
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
So, there will be two vectors of unit length perpendicular to vectorsNCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the correct option is (b).


FILL IN THE BLANKS

Q.34. The vectorNCERT Exemplar - Vectors Commerce Notes | EduRevbisects the angle between the non-collinear vectorsNCERT Exemplar - Vectors Commerce Notes | EduRevif ________.
Ans.
If vectorNCERT Exemplar - Vectors Commerce Notes | EduRev bisects the angle between non-collinear vectorsNCERT Exemplar - Vectors Commerce Notes | EduRev 
then the angle betweenNCERT Exemplar - Vectors Commerce Notes | EduRevis equal to the angle betweenNCERT Exemplar - Vectors Commerce Notes | EduRev
So,NCERT Exemplar - Vectors Commerce Notes | EduRev  ...(i)

Also,NCERT Exemplar - Vectors Commerce Notes | EduRev[∵ θ is same]   ...(ii)
From eq. (i) and eq. (ii) we get,
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required filler isNCERT Exemplar - Vectors Commerce Notes | EduRev

Q.35. IfNCERT Exemplar - Vectors Commerce Notes | EduRevfor some non-zero vectorNCERT Exemplar - Vectors Commerce Notes | EduRevthen the value of NCERT Exemplar - Vectors Commerce Notes | EduRevis   ________
Ans.
IfNCERT Exemplar - Vectors Commerce Notes | EduRevis a non-zero vector, thenNCERT Exemplar - Vectors Commerce Notes | EduRevcan be in the same plane.
Since angles between NCERT Exemplar - Vectors Commerce Notes | EduRev and  NCERT Exemplar - Vectors Commerce Notes | EduRev  are zero i.e. θ = 0
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence the required value is 0.

Q.36. The vectorsNCERT Exemplar - Vectors Commerce Notes | EduReva re the adjacent sides of a parallelogram. The acute angle between its diagonals is ________.
Ans.
Given thatNCERT Exemplar - Vectors Commerce Notes | EduRev
andNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Let θ be the angle between the two diagonal vectorsNCERT Exemplar - Vectors Commerce Notes | EduRevNCERT Exemplar - Vectors Commerce Notes | EduRev
then
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence the value of required filler isNCERT Exemplar - Vectors Commerce Notes | EduRev

Q.37. The values of k for whichNCERT Exemplar - Vectors Commerce Notes | EduRev is parallel toNCERT Exemplar - Vectors Commerce Notes | EduRevholds true are _______.
Ans.
Given thatNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Now sinceNCERT Exemplar - Vectors Commerce Notes | EduRevis parallel toNCERT Exemplar - Vectors Commerce Notes | EduRev
Here we see that atNCERT Exemplar - Vectors Commerce Notes | EduRevbecome null vector and then it will not be
parallel toNCERT Exemplar - Vectors Commerce Notes | EduRev
Now sinceNCERT Exemplar - Vectors Commerce Notes | EduRevis parallel toNCERT Exemplar - Vectors Commerce Notes | EduRev
Here we see that atNCERT Exemplar - Vectors Commerce Notes | EduRevbecome null vector and then it will not be
parallel toNCERT Exemplar - Vectors Commerce Notes | EduRev 
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the required value of k ∈ (- 1, 1) and k ≠NCERT Exemplar - Vectors Commerce Notes | EduRev

Q.38. The value of the expressionNCERT Exemplar - Vectors Commerce Notes | EduRevis _______.
Ans.
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the value of the filler isNCERT Exemplar - Vectors Commerce Notes | EduRev

Q.39. IfNCERT Exemplar - Vectors Commerce Notes | EduRevandNCERT Exemplar - Vectors Commerce Notes | EduRevis equal to _______.
Ans.
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the value of the filler is 3.

Q.40. IfNCERT Exemplar - Vectors Commerce Notes | EduRevany non-zero vector, thenNCERT Exemplar - Vectors Commerce Notes | EduRev
equals _______.
Ans.
LetNCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
= a1
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the value of the filler isNCERT Exemplar - Vectors Commerce Notes | EduRev

State True or False in each of the following Exercises.
Q.41. IfNCERT Exemplar - Vectors Commerce Notes | EduRevthen necessarily it impliesNCERT Exemplar - Vectors Commerce Notes | EduRev
Ans.
IfNCERT Exemplar - Vectors Commerce Notes | EduRevthenNCERT Exemplar - Vectors Commerce Notes | EduRevwhich is true.
Hence, the statement is True.

Q.42. Position vector of a point P is a vector whose initial point is origin.

Ans.
True

Q.43. IfNCERT Exemplar - Vectors Commerce Notes | EduRevthen the vectors
NCERT Exemplar - Vectors Commerce Notes | EduRevare orthogonal.
Ans.
Given thatNCERT Exemplar - Vectors Commerce Notes | EduRev
Squaring both sides, we get
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
NCERT Exemplar - Vectors Commerce Notes | EduRev
which implies thatNCERT Exemplar - Vectors Commerce Notes | EduRevare orthogonal.
Hence the given statement is True.

Q.44. The formulaNCERT Exemplar - Vectors Commerce Notes | EduRevis valid for non-zero vectors
NCERT Exemplar - Vectors Commerce Notes | EduRev
Ans.
NCERT Exemplar - Vectors Commerce Notes | EduRev
Hence, the given statement is False.

Q.45. IfNCERT Exemplar - Vectors Commerce Notes | EduRevare adjacent sides of a rhombus, thenNCERT Exemplar - Vectors Commerce Notes | EduRev
Ans.
IfNCERT Exemplar - Vectors Commerce Notes | EduRev
So the angle between the adjacent sides of the rhombus should be 90° which is not possible.
Hence, the given statement is False.

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