Rd Sharma XII Vol 2 2020 2021 Solutions for Class 12 Commerce Maths Chapter 6 Vector Or Cross Product are provided here with simple step-by-step explanations. These solutions for Vector Or Cross Product are extremely popular among Class 12 Commerce students for Maths Vector Or Cross Product Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2020 2021 Book of Class 12 Commerce Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2020 2021 Solutions. All Rd Sharma XII Vol 2 2020 2021 Solutions for class Class 12 Commerce Maths are prepared by experts and are 100% accurate.

Page No 24.30:

Question 1:

If a =i^+3j^-2k^ and b =-i^+3k^, find  a ×b .

Answer:

Given:a=i^+3j^-2k^ b=-i^+0j^+3k^ a×b=i^j^k^13-2-103               =9+0 i^-3-2 j^+0+3 k^                =9 i ^-j^+3k^a×b=92+-12+32                =91

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Question 2:

(i) If a =3i^+4j^ and b =i^+j^+k^, find the value of  a ×b .

(ii) If a =2i^+k^, b =i^+j^+k^, find the magnitude of a ×b .

Answer:

i Given: a=3i^+4j^+0k^ b=i^+j^+k^ a×b=i^j^k^340111         =4-0 i^ -3-0 j^ +3-4 k^          =4 i^-3j^-k^a×b=16+9+1                =26

ii Given:a=2i^+0j^+k^ b=i^+j^+k^a×b=i^j^k^201111         =0-1 i^ -2-1 j^ +2-0 k^          =- i^-j^+2k^a×b=-12+-12+22                =6

Page No 24.30:

Question 3:

(i) Find a unit vector perpendicular to both the vectors 4i^-j^+3k^ and -2i^+j^-2k^.

(ii) Find a unit vector perpendicular to the plane containing the vectors a =2i^+j^+k^ and b =i^+2j^+k^.

Answer:

i Given: a=4i^-j^+3k^  b=-2i^+j^-2k^ a×b=i^j^k^4-13-21-2          =2-3 i^ - -8+6 j^ +4-2 k^           =- i^+2j^+2k^a×b=1+22+22                =9                =3Unit vector perpendicular to a and b =a×ba×b=- i^+2j^+2k^3

ii Given:a=2i^+j^+k^ b=i^+2j^+k^ a×b=i^j^k^211121               =1-2 i^ -2-1 j^ +4-1k                 =-i^-j^+3k^a×b=1+1+9                 =11Unit vector perpendicular to the plane containing vectors a and b=±a×ba×bUnit vector perpendicular to the plane containing vectors a and b=±111-i^-j^+3k^

Page No 24.30:

Question 4:

Find the magnitude of a =3k^+4j^×i^+j^-k^.

Answer:

a=0i^+4j^+3k^×i^+j^-k^   =i^j^k^04311-1  =i^ -4-3-j^ 0-3+k^ 0-4  =-7i^+3j^-4k^a=-72+32+-42          =74

Page No 24.30:

Question 5:

If a =4i^+3j^+k^ and b =i^-2k^, then find  2b^×a .

Answer:

Given:a=4i^+3j^+k^ 2b=2i^+0j^-4k^2b×a=i^j^k^20-4431           =0+12 i^-2+16 j^+6-0k^            =12i^-18j^+6k^2b×a=122+-182+62                   =504

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Question 6:

If a =3i^-j^-2k^ and b =2i^+3j^+k^, find  a +2b × 2a -b .

Answer:

Given:a=3i^-j^-2k^ b=2i^+3j^+k^ a+2b=3i^-j^-2k^+2 2i^+3j^+k^                 =7i^+5j^+0k^ 2a-b=2 3i^-j^-2k^- 2i^+3j^+k^                 =4i^-5j^-5k^a+2b×2a-b=i^j^k^7504-5-5                                 =i^ -25+0-j^ -35+0+k^ -35-20                                 =-25i^+35j^-55k^

Page No 24.30:

Question 7:

(i) Find a vector of magnitude 49, which is perpendicular to both the vectors 2i^+3j^+6k^ and 3i^-6j^+2k^.

(ii) Find a vector whose length is 3 and which is perpendicular to the vector a =3i^+j^-4k^ and b =6i^+5j^-2k^.

Answer:

i Given:a=2i^+3j^+6k^ b=3i^-6j^+2k^a×b=i^j^k^2363-62          =6+36 i^ - 4-18 j^ +-12-9 k^            =42i^+14j^-21k^a×b=422+142+-212                =2401                =49Required vector =49×a×ba×b                              =49 ×42i^+14j^-21k^49                               =42i^+14j^-21k^

ii Given:a=3i^+j^-4k^ b=6i^+5j^-2k^a ×b=i^j^k^31-465-2               = -2+20 i^--6+24 j^ +15-6 k^                =18i^-18j^+9k^a×b=182+-182+92                =729                =27Required vector=3×a×ba×b                            =3×18i^-18j^+9k^27                             =32i^-2j^+k^3                             =2i^-2j^+k^

Page No 24.30:

Question 8:

Find the area of the parallelogram determined by the vectors:
(i) 2i^ and 3j^

(ii) 2i^+j^+3k^ and i^-j^

(iii) 3i^+j^-2k^ and i^-3j^+4k^

(iv) i^-3j^+k^ and i^+j^+k^.

Answer:

i Let:a=2i^+0j^+0k^ b=0i^+3j^+0k^ a×b=i^j^k^200030              =0-0 i^ -0-0 j^ + 6-0 k^              =0 i^+0j^+6k^Area of the parallelogram=a×b                                                                                                      =0+0+62                                              =6 sq. units

ii Let: a =2i^+j^+3k^ b=i^-j^+0k^ a×b=i^j^k^2131-10              =0+3 i^ -0-3 j^ +-2-1 k^               = 3i^+3j^-3k^Area of the parallelogram =a×b                                             =32+32+32                                             =27                                             =33 sq. units

iii Let:a=3i^+j^-2k^b=1i^-3j^+4k^a×b=i^j^k^31-21-34         =i^ 4-6-j^ 12+2+k^ -9-1         =-2i^-14j^-10k^Area of the parallelogram=a×b                                             =-22+-142+-102                                             =300                                             =103 sq. units

iv Let: a=i^-3j^+k^ b=i^+j^+k^a×b=i^j^k^1-31111         =-3-1 i^ - 1-1 j^+1+3 k^          =-4i^+0j^+4k^Area of the parallelogram=a×b                                            =-42+0+42                                            =32                                            =42 sq. units.

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Question 9:

Find the area of the parallelogram whose diagonals are:
(i) 4i^-j^-3k^ and -2j^+j^-2k^

(ii) 2i^+k^ and i^+j^+k^

(iii) 3i^+4j^ and i^+j^+k^

(iv) 2i^+3j^+6k^ and 3i^-6j^+2k^

Answer:

i Let:a=4i^-j^-3k^b=-2i^+j^-2k^ a×b=i^j^k^4-1-3-21-2              =2+3 i^ --8-6j^ + 4-2k^              =5i^+14j^+2k^a×b=25+196+4                =225                =15Area of the parallelogram = 12a×b                                              =152 sq. units.

ii Let: a=2i^+0j^+k^ b=i^+j^+k^a×b=i^j^k^201111         =0-1 i^ -2-1 j^ +2-0 k^          =-i^-j^+2k^a×b=-12+-12+2                =6Area of the parallelogram = 12a×b                                              =62 sq. units

iii Let:a=3i^+4j^+0k^ b=i^+j^+k^ a×b=i^j^k^340111              =4-0 i^ -3-0 j^ +3-4 k^               =4i^-3j^-k^a×b=42+-32+-12                =26Area of the parallelogram = 12a×b                                             =262sq. units

iv Let:a=2i^+3j^+6k^b=3i^-6j^+2k^ a×b=i^j^k^2363-62              = 6+36i^-4-18j^ +-12-9k^               =42i^+14j^-21k^a×b=422+142+-212                 =2401                 =49Area of the parallelogram = 12a×b                                             =492 sq. units

Disclaimer: The answer given for (iii) and (iv) in the textbook is incorrect.

Page No 24.30:

Question 10:

If a =2i^+5j^-7k^, b =-3i^+4j^+k^ and c =i^-2j^-3k^, compute  a ×b ×c and a × b ×c  and verify that these are not equal.

Answer:

Given: a=2i^+5j^-7k^ b=-3i^+4j^+k^ c=i^-2j^-3k^ a×b=i^j^k^25-7-341              =5+28 i^ -2-21 j^ +8+15k^               =33i^+19j^+23k^a×b×c=i^j^k^3319231-2-3                        = -57+46i^- -99-23 j^+ -66-19 k^a×b×c =-11i^+122j^-85k^         ...(1) b×c=i^j^k^-3411-2-3               = -12+2 i^- 9-1 j^+6-4 k^               =-10i^-8j^+2k^ a×b×c=i^j^k^25-7-10-82                         =10-56 i^ -4-70 j^ +-16+50 k^  a×b×c=-46i^+66j^+34k^          ...(2)From (1) and (2), we geta×b×ca×b×c

Page No 24.30:

Question 11:

If  a =2, b =5 and  a ×b =8, find a ·b .

Answer:

We know  a.b2+a ×b2=a2b2a.b2+82=22×52          ( a ×b=8, a=2 and b=5)a.b2+64=100a.b2=36a.b=6



Page No 24.31:

Question 12:

Given a =172i^+3j^+6k^, b =173i^-6j^+2k^, c =176i^+2j^-3k^, i^, j^, k^ being a right handed orthogonal system of unit vectors in space, show that a , b , c is also another system.

Answer:

Given: a=17 2i^+3j^+6k^ b=17 3i^-6j^+2k^ c=176 i^+2j^-3k^a×b=17 17i^j^k^2363-62        =14942 i^+14j^-21k^        =1497 6 i^+2j^-3k^        =176 i^+2j^-3k^        =cb×c=17 17i^j^k^3-6262-3        =14914 i^+21j^+42k^        =1497 2i^+3j^+6k^        =17 2i^+3j^+6k^        =ac×a=17 17i^j^k^62-3236        =14921 i^-42j^+14k^        =1497 3i^-6j^+2k^        =17 3i^-6j^+2k^        =ba=174+9+36   =77   =1b=179+36+4    =77    =1c=1736+4+9   =77   =1Thus, ab and c form a right handed orthogonal system of unit vectors.

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Question 13:

If  a =13, b =5 and a . b =60, then find  a ×b .

Answer:

We know a.b2+a ×b2=a2b2602+a ×b2=132×52          ( a.b=60, a=13 and b =5)3600+a ×b2=4225a ×b2=625a ×b=25

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Question 14:

Find the angle between two vectors a and b , if  a ×b =a ·b .

Answer:

Let θ be the angle between aand b.Given:a×b=a. ba b sin θ=a b cos θsin θ=cos θtan θ = 1θ=π4

Page No 24.31:

Question 15:

If a ×b =b ×c 0, then show that a +c =mb, where m is any scalar.

Answer:

a×b=b×ca×b=-c×ba×b+c×b=0a+c×b=0           (Using right distributive property)Thus, a+c is parallel to b.a+c=mb, for some scalar m.

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Question 16:

If  a =2, b =7 and a ×b =3i^+2j^+6k^, find the angle between a and b .

Answer:

Let θ be the angle between a and b.a×b=3i^+2j^+6k^    (Given)a×b=9+4+36                =7We know a×b=a b sin θ7=2 7 sin θsin θ=12θ=π6

Page No 24.31:

Question 17:

What inference can you draw if a ×b =0  and a ·b =0.

Answer:

Given:a×b=0a=0      b=0  ab     Also,a.b=0a b cos θ=0 a=0 or b=0 or, ab      But a cannot be both perpendicular as well as parallel to b. a=0      b=0

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Question 18:

If a, b, c are three unit vectors such that a ×b =c , b ×c =a, c ×a =b . Show that a, b, c form an orthonormal right handed triad of unit vectors.

Answer:

Given:a×b=cb×c=ac×a=b          ...(1)Now,a×b=c=1      (∵ c is a unit vector)b×c=a=1      (∵ a is a unit vector)c×a=b=1      (∵ b is a unit vector) a×b=b×c=c×a=1        ...(2)From (1) and (2), we know a, b and c form an orthonormal right handed triad of unit vectors.

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Question 19:

Find a unit vector perpendicular to the plane ABC, where the coordinates of A, B and C are A (3, −1, 2), B (1, −1, −3) and C (4, −3, 1).

Answer:

The vector AB×AC is perpendicular to the vectors AB and AC. Required unit vector =AB×ACAB×ACNow,AB =Position vector of B- Position vector of A       =i^-j^-3k^-3i^-j^+2k^       =-2i^+0j^-5kAC  =Position vector of C- Position vector of A        =4i^-3j^+k^-3i^-j^+2k^        =i^-2j^-k^ AB×AC=i^j^k^-20-51-2-1                   =0-10 i-2+5 j+4-0 k^                   =-10i^-7j^+4k^AB×BC=-102+-72+42                 =165Unit vector perpendicular to the plane ABC =AB×ACAB×AC=-10i^-7j^+4k^165

Page No 24.31:

Question 20:

If a, b, c are the lengths of sides, BC, CA and AB of a triangle ABC, prove that BC+CA+AB=0 and deduce that asin A=bsin B=csin C.

Answer:

We haveBC=a CA=bAB=ca=ab=b       ( Length is always positive)c=c  Now,   BC+CA+AB=0           (Given)a+b+c=0a×a+b+c=a×0a×a+a×b+a×c=00+a×b-c×a=0a×b=c×aa bsin C=c a sin Bab sin C=ca sin BDividing both sides by abc, we getsin Cc=sin Bb        ...(1)Again,BC+CA+AB=0a+b+c=0b×a+b+c=b×0b×a+b×b+b×c=0-a×b+0+b×c=0a×b=b×ca b sin C=b c sin Aab sin C=bc sin ADividing both sides by abc, we getsin Cc=sin Aa        ...(2)From (1) and (2), we getsin Aa=sin Bb=sin Ccasin A=bsin B=csin C

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Question 21:

If a =i^-2j^+3k^, and b =2i^+3j^-5k^, then find a ×b . Verify that a and a ×b are perpendicular to each other.

Answer:

Given:a=i-2j+3k b=2i+3j-5k

a×b=i^j^k^1-2323-5         =i^+11j^+7k^Now, a. a×b=1-22+21                 =0Thus, a is perpendicular to a×b.

Page No 24.31:

Question 22:

If p and q are unit vectors forming an angle of 30°; find the area of the parallelogram having a =p +2q and b =2p +q as its diagonals.

Answer:

Given: a=p^+2q b= 2p+q^a×b=p+2q×2p+q=2 p×p+p×q+4q×p+2 q×q=20+p×q-4 p×q+2 0=-3p×qArea of the parallelogram=12a×b                                             =12-3p×q                                             =32p q sin 30o                                             =32 1 1 12         (∵ p and q are unit vectors)                                             =34sq. units

Page No 24.31:

Question 23:

For any two vectors a and b , prove that  a ×b 2=a .a a .b b .a b .b .

Answer:

RHS=a. aa. bb. ab. b         =a2a b cos θa b cos θb2        =a2 b2-a2 b2 cos2 θ        =a2 b2 1-cos2 θ        =a2 b2  sin2 θ        =ab sin θ2        =a×b2        =LHSHence proved.

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Question 24:

Define a ×b and prove that  a ×b = a .b  tan θ, where θ is the angle between a and b .

Answer:

If a and b are two non-zero non-parallel vectors, then the vector product denoted by a×b is defined as a×b=a b sin θ η^.Here, θ is the angle between a and b and η^ is the unit vector perpendicular to the plane of a and b such that a, b and η^ form a right handed system.LHS=a×b        =a b sin θ         =a b sin θ ×cos θcos θ        =a b cos θ sin θcos θ        =a b cos θ tan θ        =a. b tan θ        =RHSHence proved.

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Question 25:

If  a =26, b =7 and  a ×b =35, find a .b .

Answer:

We knowa.b2+a ×b2=a2b2a.b2+352=262×72          ( a×b=35, a=26 and b =7)a.b2+1225=1274a.b2=49a.b=7

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Question 26:

Find the area of the triangle formed by O, A, B when OA=i^+2j^+3k^, OB=-3i^-2j^+k^.

Answer:

Given:OA =i^+2j^+3k^OB=-3i^-2j^+k^ OA×OB=i^j^k^123-3-21               =8i^-10j^+4k^OA×OB=64+100+16                      =180                      =65Area of the triangle=12OA×OB                                 =12 65                                 =3 5 sq. units

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Question 27:

(i) Let a =i^+4j^+2k^, b =3i^-2j^+7k^ and c =2i^-j^+4k^. Find a vector d which is perpendicular to both a and b and c ·d =15.

(ii) Let a=4i^+5j^-k^, b=i^-4j^+5k^ and c=3i^+j^-k^. Find a vector d   which is perpendicular to both c and b and d.a=21.

Answer:

(i)
Given:a=i^+4j^+2k^ b=3i^-2j^+7k^c=2i^-j^+4k^Since d is perpendicular to both a and b, it is parallel to a×b.Suppose d=λa×b for some scalar λ.d=λ i^j^k^1423-27   =λ 28+4i^-7-6 j^+-2-12k^   =λ 32i^- j^-14k^c.d=15              (Given)2i^-j^+4k^.λ 32i^- j^-14k^=15λ64+1-56=15λ=53 d=5332i^- j^-14k^d=13160i^- 5j^-70k^

Disclaimer: The question should contain
"which is perpendicular to both a and binstead of "which is perpendicular to both a and d

(ii)

Let d=x i^+y j^ +z k^ Since d is perpendicular to both c and b , sod.c =0 and d.b =03x+y-z=0                 .....1x-4y+5z= 0              .....2d.a =214x+5y-z=21             .....3Solving 1, 2 and 3x=-13, y=163, z=133d=13- i^+16 j^ +13 k^d=-13 i^-16 j^ -13 k^

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Question 28:

Find a unit vector perpendicular to each of the vectors a +b and a -b , where a =3i^+2j^+2k^ and b =i^+2j^-2k^.

Answer:

Given: a=3i^+2j^+2k^b=i^+2j^-2k^ a+b=4i^+4j^+0k^     a-b=2i^+0j^+4k^a+b×a-b=i^j^k^440204                            =16i^-16j^-8k^ a+b×a-b=256+256+64                                  =576                                  =24Unit vector perpendicular to both a+b and a-b=a+b×a-ba+b×a-b=16i^-16j^-8k^24=8 2i^-2j^-k^24=132i^-2j^-k^

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Question 29:

Using vectors find the area of the triangle with vertices, A (2, 3, 5), B (3, 5, 8) and C (2, 7, 8).

Answer:

Let a, b and c be the position vectors of A, B and C, respectively. Then,a=2i^+3j^+5k^b=3i^+5j^+8k^c=2i^+7j^+8k^Now, AB=b-a     =i^+2j^+3k^AC=c-a      =0i^+4j^+3k^ AB×AC=i^j^k^123043                    =-6i^-3j^+4k^AB×AC=36+9+16                      =61Area of triangle ABC=12AB×AC                                    =612sq. units

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Question 30:

If a=2i^-3j^+k^, b=-i^+k^, c=2j^-k^ are three vectors, find the area of the parallelogram having diagonals a+b and b+c.     [CBSE 2014]

Answer:


It is given that a=2i^-3j^+k^, b=-i^+k^, c=2j^-k^.

a+b=2i^-3j^+k^+-i^+k^=i^-3j^+2k^

 b+c=-i^+k^+2j^-k^=-i^+2j^

We know that the area of parallelogram is 12d1×d2, where d1 and d2 are the diagonal vectors.

Now,

a+b×b+c=i^j^k^1-32-120=-4i^-2j^-k^

∴ Area of the parallelogram having diagonals a+b and b+c

 =12a+b×b+c=12-4i^-2j^-k^=12-42+-22+-12=212 square units

Thus, the required area of the parallelogram is 212 square units.

Page No 24.31:

Question 31:

The two adjacent sides of a parallelogram are 2i^-4j^+5k^ and i^-2j^-3k^. Find the unit vector parallel to one of its diagonals. Also, find its area.

Answer:

Suppose  ABCD is the given parallelogram and AC is its diagonal.Let:AB=2i^-4j^+5k^BC=i^-2j^-3k^ Diagonal AC=AB+BC                        =3i^-6j^+2k^AC=9+36+4            =7Unit vector parallel to AC=ACAC                                            =3i^-6j^+2k^7Now,AB×BC=i^j^k^2-451-2-3               =22i^+11j^+0k^AB×AC=484+121                      =605                      =115Area of triangle ABC=12AB×AC                                    =1152sq. units



Page No 24.32:

Question 32:

If either a =0 or b =0 , then a ×b =0 . Is the converse true? Justify your answer with an example.

Answer:

If a=0 or b=0, then a b sin θ n^=0.a×b=0But the converse is not true as whenever a×b=0, we cannot be sure that either a=0 or b=0.For example:a=i^+2j^+3k^b=i^+2j^+3k^Here,a≠0b≠0But a×b=i^j^k^123123                  =0i^+0j^+0k^                  =0

Page No 24.32:

Question 33:

If a =a1i^+a2j^+a3k^, b =b1i^+b2j^+b3k^ and c =c1i^+c2j^+c3k^, then verify that a × b +c =a ×b +a ×c .

Answer:

Given:a=a1i^+a2j^+a3k^b=b1i^+b2j^+b3k^ c=c1i^+c2j^+c3k^b+c=b1+c1 i^+b2+c2 j^+b3+c3 k^ a×b+c=i^j^k^a1a2a3b1+c1b2+c2b3+c3                       =a2b3+a2c3-a3b2-a3c2i^-a1b3+a1c3-a3b1-a3c1j^+ a1b2+a1c2-a2b1-a2c1k^        ...(1)Now, a×b=i^j^k^a1a2a3b1b2b3          = a2b3-a3b2i^- a1b3-a3b1j^+ a1b2-a2b1k^a×c=i^j^k^a1a2a3c1c2c3          =a2c3-a3c2i^ -a1c3-a3c1j^ +a1c2-a2c1k^ a×b+b×c= a2b3+a2c3-a3b2-a3c2i^-a1b3+a1c3-a3b1-a3c1 j^ + a1b2+a1c2-a2b1-a2c1k^      ...(2)From (1) and (2), we geta×b+c=a×b+b×c

Page No 24.32:

Question 34:

Using vectors, find the area of the triangle with vertices:
(i) A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5)
(ii) A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1)                                   [CBSE 2011, NCERT EXEMPLAR]
                                                 

Answer:


(i) The vertices of the triangle are A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Position vector of A = i^+j^+2k^

Position vector of B = 2i^+3j^+5k^

Position vector of C = i^+5j^+5k^

AB=2i^+3j^+5k^-i^+j^+2k^=i^+2j^+3k^

AC=i^+5j^+5k^-i^+j^+2k^=4j^+3k^

Now,

AB×AC=i^j^k^123043=-6i^-3j^+4k^

∴ Area of ∆ABC = 12AB×AC

                          =12-6i^-3j^+4k^=12-62+-32+42=1236+9+16=612 square units

(ii) The vertices of the triangle are A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1).

Position vector of A = i^+2j^+3k^

Position vector of B = 2i^-j^+4k^

Position vector of C = 4i^+5j^-k^

AB=2i^-j^+4k^-i^+2j^+3k^=i^-3j^+k^

AC=4i^+5j^-k^-i^+2j^+3k^=3i^+3j^-4k^

Now,

AB×AC=i^j^k^1-3133-4=9i^+7j^+12k^

∴ Area of ∆ABC = 12AB×AC

                          =129i^+7j^+12k^=1292+72+122=1281+49+144=2742 square units

Page No 24.32:

Question 35:

Find all vectors of magnitude 103 that are perpendicular to the plane of i^+2j^+k^ and -i^+3j^+4k^.                  [NCERT EXEMPLAR]

Answer:


Let a=i^+2j^+k^ and b=-i^+3j^+4k^.

Unit vectors perpendicular to both a and b = ±a×ba×b

Now,
a×b=i^j^k^121-134=5i^-5j^+5k^a×b=5i^-5j^+5k^=52+-52+52=75=53

Unit vectors perpendicular to both a and b = ±5i^-5j^+5k^53=±i^-j^+k^3

∴ Required vectors = 103±i^-j^+k^3=±10i^-j^+k^

Thus, the vectors of magnitude 103 that are perpendicular to the plane of i^+2j^+k^ and -i^+3j^+4k^ are ±10i^-j^+k^.

Page No 24.32:

Question 36:

The two adjacent sides of a parallelogram are 2i^-4j^-5k^ and 2i^+2j^+3k^. Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.

Answer:

The two adjacent sides of a parallelogram are 2i^-4j^-5k^ and 2i^+2j^+3k^.
Suppose a=2i^-4j^-5k^ and  b=2i^+2j^+3k^



Then any one diagonal of a parallelogram is  P=a+b.
P=a+b=2i^-4j^-5k^ +2i^+2j^+3k^=4i^-2j^-2k^
Therefore, unit vector along the diagonal is PP=4i^-2j^-2k^16+4+4=2i^-j^-k^6.
Another diagonal of a parallelogram is P'=b-a.
P'=b-a=2i^+2j^+3k^-2i^+4j^+5k^=6j^+8k^
Therefore, unit vector along the diagonal is P'P'=6j^+8k^36+64=6j^+8k^10=3j^+4k^5.
Now, 
P×P'=i^j^k^4-2-2068=i^-16+12-j^32-0+k^24-0=-4i^-32j^+24k^
Area of parallelogram = P×P'2=16+1024+5762=16162=41012=2101 square units

Page No 24.32:

Question 37:

If a×b2+a·b2=400 and a=5, then write the value of b.

Answer:

a×b2+a·b2=400absinθ2+abcosθ2=400a2b2sin2θ+a2b2cos2θ=400a2b2=40025×b2=400b2=16b=4

Page No 24.32:

Question 38:

If θ is the angle between two vectors i^-2j^+3k^ and 3i^-2j^+k^, find sin θ.

Answer:

Let a=i^-2j^+3k^ and b=3i^-2j^+k^. If θ is the angle between them. Then,
cosθ=a·bab
Now,
a·b=(i^-2j^+3k^)·(3i^-2j^+k^)        =3+4+3=10

a=1+4+9=14 and b=9+4+1=14
cosθ=a·babcosθ=101414=1014=57
Now,
sinθ=1-cos2θ        =1-572        =2449=267



Page No 24.34:

Question 1:

If a is any vector, then a ×i^2+a ×j^2+a ×k^2=
(a) a2

(b) 2a2

(c) 3a2

(d) 4a2

Answer:

(b) 2a2

Let a=a1 i^+a2j^+a3 k^a×i^=i^j^k^a1a2a3100        =a3j^-a2k^ a×i^2=a3j^-a2k^ 2                 =a32j^2+a22k^2-2 a3a2 j^. k^                 =a32+a22                (∵ j^. k^=0)          ...(1) a×j^=i^j^k^a1a2a3010              =-a3i^+a1k^a×j^2=-a3i^+a1k^ 2                  =a32i^2+a12k^2-2 a3a2 i^. k^                  =a32+a12             (∵ i^. k^=0)         ...(2)  a×k^=i^j^k^a1a2a3001              =a2i^-a1j^a×k2=a2i^-a1j^2                  =a22i^2+a12j2+2 a1a2 i^. j^                  =a22+a12             (∵ i^. j^=0)      ...(3)Adding (1), (2) and (3), we geta×i^2+a×j^2+a×k2=a32+a22+a32+a12+a22+a12                                               =2 a12+a22+a32                                              =2 a2                    (∵ a=a12+a22+a32 ) 

Page No 24.34:

Question 2:

If a ·b =a ·c and a ×b =a ×c, a 0, then
(a) b =c 

(b) b =0 

(c) b +c =0 

(d) none of these

Answer:

(a) b =c 

a ·b =a ·c  a  ·b -a ·c =0 a .b -c =0   Let θ be the angle between a and b-c     ab -c cos θ      ...(1)

and  a ×b =a ×c  a ×b -a ×c =0a  ×b -c =0Then , a b -c  sin θ=0      ...(2)Here, it is given that a 0Therefore, for eq (1) and eq (2) to be 0We have , b -c  cos θ=0 For   b -c  cos θ=0 , one of   b -c  or cos θ must be 0Case 1:Let cos  θ=0θ=90°sin θ=1& if  b -c  sin θ=0 and sin θ=1 Then  b -c =0b =cCase 2:Let b -c =0b =cHence, b =c 

Page No 24.34:

Question 3:

The vector b =3i^+4k^ is to be written as the sum of a vector α parallel to a =i^+j^ and a vector β perpendicular to a . Then α =
(a) 32i^+j^

(b) 23i^+j^

(c) 12i^+j^

(d) 13i^+j^

Answer:

(a) 32i^+j^

Let:α=a1i^+a2j^+a3k^β=b1i^+b2j^+b3k^Now,b=3i^+4k^=α+β                                           (Given)3i^+0j^+4k^=a1+b1 i^+a2+b2 j^+a3+b3 k^a1+b1=3; a2+b2=0; a3+b3=4a1+b1=3; a2=-b2; a3+b3=4              ...(1)a=i^+j^                                                           (Given) Also, α is parallel to a.α×a=0i^j^k^a1a2a3110=0-a3i^+a3j^+a1-a2k^=0i^+0j^+0k^a3=0; a1-a2=0 a3=0; a1=a2                             ...(2)Since β is perpendicular to a, we getβ. a=0b1i^+b2j^+b3k^. i^+j^=0b1+b2=0b1=-b2                                             ...(3)Solving (1), (2) and (3), we geta1=32; a2=32; a3=0∴ α=a1i^+a2j^+a3k^      =32i^+32j^+0k^      =32 i^+j^

Page No 24.34:

Question 4:

The unit vector perpendicular to the plane passing through points Pi^-j^+2k^, Q2i^-k^ and R2j^+k^ is
(a) 2i^+j^+k^

(b) 62i^+j^+k^

(c) 162i^+j^+k^

(d) 162i^+j^+k^

Answer:

(c) 162i^+j^+k^


The vector  PQ×PR is perpendicular to the vectors PQ  and PR. Required unit vector==PQ×PRPQ×PRNow, PQ=P.V. of Q-P.V. of P                  =i^+j^-3k^PR=P.V. of R-P.V. of P      =-i^+3j^-k^ PQ×PR=i^j^k^11-3-13-1                    =8i^+4j^+4k^                    =4 2i^+j^+k^PQ×PR=64+16+16                      =96                     =46Required unit vector=PQ×PRPQ×PR                                   =4 2i^+j^+k^46                                   = 162i^+j^+k^

Page No 24.34:

Question 5:

If a, b represent the diagonals of a rhombus, then
(a) a ×b =0 

(b) a ·b =0

(c) a ·b =1

(d) a ×b =a 

Answer:

(b) a ·b =0

We know that the diagonals in a rhombus (a and b) are perpendicular.Therefore, their dot product is zero.a. b=0

Page No 24.34:

Question 6:

Vectors a and b are inclined at angle θ = 120°. If  a =1, b =2, then a +3b ×3a -b 2 is equal to
(a) 300
(b) 325
(c) 275
(d) 225

Answer:

(a) 300

a+3b×3a-b=3 a×a-a×b+9 b×a-3 b×b=3 0-a×b-9 a×b-3 0=-10 a×bNow,a×3b×3a-b 2=-10 a×b2=100 a×b2=100 a2 b2 sin2120=100 12 22 322=400 ×34=300

Page No 24.34:

Question 7:

If a =i^+j^-k^, b =-i^+2j^+2k^ and c =-i^+2j^-k^, then a unit vector normal to the vectors a +b and b -c  is
(a) i^

(b) j^

(c) k^

(d) none of these

Answer:

(a) i^

a+b=0i^+3j^+k^b-c=0i^-0j^+3k^a+b×b-c=i^j^k^031003                            =9i^a+b×b-c=9  i^                              =91                              =9Unit vector perpendicular to both a+b and b-c=a+b×b-ca+b×b-c=9i^9=i^

Page No 24.34:

Question 8:

A unit vector perpendicular to both i^+j^ and j^+k^ is

(a) i^-j^+k^

(b) i^+j^+k^

(c) 13i^+j^+k^

(d) 13i^-j^+k^

Answer:

(d) 13i^-j^+k^

Let:a=i^+j^+0k^ b=0i^+j^+k^  a×b=i^j^k^110011              = i^-j^+k^a×b=1+1+1                =3Unit vector perpendicular to a and b=a×ba×b=i^-j^+k^3


Disclaimer: The answer given for this question in the textbook is incorrect.

Page No 24.34:

Question 9:

If a =2i^-3j^-k^ and b =i^+4j^-2k^, then a ×b is
(a) 10i^+2j^+11k^

(b) 10i^+3j^+11k^

(c) 10i^-3j^+11k^

(d) 10i^-2j^-10k^

Answer:

(b) 10i^+3j^+11k^

a×b=i^j^k^2-3-114-2         =10 i ^+3j^+11k^

Page No 24.34:

Question 10:

If i^, j^, k^ are unit vectors, then
(a) i^·j^=1

(b) i^·i^=1

(c) i^×j^=1

(d) i^×j^×k^=1

Answer:

(b)  i^·i^=1

Let us check each option one by one.(a) We knowi^. j^=0    1(b) We knowi^. i^=i^2     =12     =1(c) i^×j^ =k^              1(d) i^ ×j^×k^= i^ ×i^                      =0                      1

Page No 24.34:

Question 11:

If θ is the angle between the vectors 2i^-2j^+4k^ and 3i^+j^+2k^, then sin θ =
(a) 23

(b) 27

(c) 27

(d) 27

Answer:

(b) 27

Let:a=2i^-2j^+4k^b=3i^+j^+2k^a=22+-22+42         =4+4+16         =24         =26 b=32+12+22        =9+1+4        =14a×b=i^j^k^2-24312          =-8i^+8j^+8k^a×b=64+64+64           =192           =8 3Let θ be the angle between a and b.a×b=a b sin θ 8 3=2614 sin θ sin θ= 8 3421               =27θ=sin-127



Page No 24.35:

Question 12:

If  a ×b =4, a ·b =2, then  a 2 b 2=

(a) 6
(b) 2
(c) 20
(d) 8

Answer:

(c) 20

We know a. b2+a×b2=a2 b2           ...(1)a. b=2                                         (Given)a. b2=a. b2From (1), we get22+42=a2 b2 a2 b2 =20

Page No 24.35:

Question 13:

The value of  a ×b 2 is

(a)  a 2+ b 2- a ·b 2

(b)  a 2  b 2- a ·b 2

(c)  a 2+ b 2-2 a ·b 

(d)  a 2+ b 2-a ·b 

Answer:

(b)  a 2  b 2- a ·b 2

a. b2+a×b2=a b cos θ2+a b sin θ2=a2 b2 cos2 θ+sin2 θ=a2 b2 1=a2 b2  a×b2=a2 b2 -a. b2Thus, the value of a×b2 is a2 b2 -a. b2.

Page No 24.35:

Question 14:

The value of i^·j^×k^+j^·i^×k^+k^·i^×j^, is
(a) 0
(b) −1
(c) 1
(d) 3

Answer:

 (c) 1

i^. j^×k^+j^. i^×k^+k^. i^×j^=i^. i^+j^. -j^+k^. k^=i^2-j^2+k^2=1-1+1=1

Page No 24.35:

Question 15:

If θ is the angle between any two vectors a and b , then  a · b = a ×b  when θ is equal to

(a) 0
(b) π/4
(c) π/2
(d) π

Answer:

(b)  π/4

Let θ be the angle between aand b.We know a×b=a b sin θa. b=a b cos θa. b=a b cos θ=a' b cos θGiven: a. b=a×ba b cos θ=a b sin θcos θ=sin θθ=π4

Page No 24.35:

Question 16:

If a=10,b=2 and a·b=12, then the value of a×b is
(a) 5               
(b) 10             
(c) 14             
(d) 16

Answer:

Given a=10, b=2a.b=12

Since a×b2=a2 b2-a.b2               By lagrange's identity=102 22-122=100×4-144=400-144
i.e a×b2=256i.e a×b=16
Hence, the correct answer is option D.

Page No 24.35:

Question 17:

The number of vectors of unit length perpendicular to the vectors a=2i^+j^+2k^ and b=j^+k^ is
(a) one             
(b) two             
(c) three           
(d) infinite

Answer:

Unit vector perpendicular to a and b is ±a×ba×b for a=2i^+j^+2k^ and b=j^+k^a×b=i^j^k^212011a×b=i^1-2 -j^2-0+k^2-0

 a×b=-12+22+22=9=3
 unit vector perpendicular to both a and b is ±-i^-2j^+2k^3=±-i^3-2j^3+2k^3

i.e. number of vectors of unit length perpendicular to the vectors a and b is two.

Hence, the correct answer is option B.

Page No 24.35:

Question 18:

If a=8,b=3 and a×b=12, then the value of a·b is

(a) 63                (b) 83               (c) 123              (d) none of these

Answer:

Given a=8           b=3and a×b=12using Lagrange's identity,a×b2=a2 b2-a.b2i.e. 122=82 32-a.b2i.e. a.b2=64×9-144i.e. a.b2=432i.e. a.b=123
Hence, the correct answer is option C.

Page No 24.35:

Question 19:

The vectors from origin O to the points A and B are a=2i^-3j^+2k^ and b=2i^+3j^+k^ respectively, then area of triangle OAB is
(a) 340              
(b) 25                (c) 229                 (d) 12229

Answer:

Area of Δ OAB=12a×bSince a×b=i^j^k^2-32231=i^-3-6-j^2-4+k^6+6

 a×b =i^-9+2j^+12k^i.e. a×b=81+4+144a×b=229 Area of Δ OAB=12229
Hence, the correct answer is option D.

Page No 24.35:

Question 1:

The value of the expression a×b2+a·b2 is ______________.

Answer:

By lagrange's identity, 

a×b2+a.b2=a2 b2

Page No 24.35:

Question 2:

If a×b2+a·b2 =144 and a=4, then b is equal to ____________.

Answer:

If given a×b2+a.b2=144 and a=4By lagrange's identity,a×b2+a.b2=a2b2i.e 144=42b2i.e. b2=14416=9i.e. b=3

Page No 24.35:

Question 3:

If a and b are unit vectors such that a×b is also a unit vector, then the angle between a and b is ___________.

Answer:

Given for a, ba, b and a×b are unit vectors.i.e a=1=b=a×b By lagrange's identity, a×b2+a.b2=a2b2i.e a.b2=1-1=0i.e a.b2=0i.e a.b=0i.e. a.b=0i.e. abcosθ=0; where θ is angle between a and bi.e. 1×1 cosθ=0i.e. cosθ=0i.e. θ=π2i.e. angle between a and b is π2

Page No 24.35:

Question 4:

For any two vectors a and b, 2a+3b×5a+7b = _________________.

Answer:

For any two vectors, a and b2a+3b×5a+7b=2a×5a+3b×5a+2a×7b+3b×7b=10a×a+15b×a+14a×b+21b×bSince a×a=0=b×b and a×b=-b×a                =0+15b×a-14b×a+02a+3b×5a+7b=b×a

Page No 24.35:

Question 5:

The number of vectors of unit length perpendicular to vectors a=i^+j^ and b=j^+k^ is _________________.

Answer:

The unit vector perpendicular to a and b is given by ±a×ba×b

for a=i^+j^ and b=j^+k^a×b=i^j^k^110011=i^1-0-j^1-0+k^1-0
a×b=i^-j^+k^i.e. a×b=1+1+1=3 number of unit vectors perpendicular to a and b is two i.e ±i^-j^+k^3



Page No 24.36:

Question 6:

If a=i^-j^ and b=j^+k^, then a×b2+a.b2= ________________.

Answer:

For a=i^-j^       b=j^+k^a=12+12=2b=12+12=2By lagrange's identity,

a×b2+a.b2=a2b2=22 22=2×2

i.e a×b2+a.b2=4

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Question 7:

For any non-zero vector r,r×i^2+r×j^2+r×k^2=λr2, then λ= ___________________.

Answer:

Let r=xi^+yj^+zk^Then r×i^=xi^+yj^+zk^×i^                 =xi^×i^+yj^×i^+zk^×i^i.e r×i^ =0-yk^+zj^           ...1    j^×i^=-k^ and k^×i^=j^Similarly,r×k^ and r×j^ are calculated i.e. r×j^=xi^+yj^+zk^×j^               =xi^×j^+yj^×j^+zk^×j^               =xk^+y0-zi^i.e. r×j^=xk^-zi^            ...2and r×k^=xi^+yj^+zk^×k^                 =xi^×k^+yj^×k^+zk^×k^                 =-xj^+yi^+z0i.e. r×k^ =-xj^+yi^         ...3 From 1, 2 and 3r×i^2+r×j^2+r×k^2=-yk^+zj^2+xk^-zi^2+-xj^+yi^2=y2+z2+x2+z2+x2+y2=2x2+y2+z2=2r2=λr2    given λ=2

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Question 8:

If a, b and c are the position vectors of the vertices A, B and C respectively of a ABC such that a×b+b×c+c×a=20,  then area of ABC is ____________.

Answer:

For a, b and c position vectors of A, B and C vertices of a ΔABC,given a×b+b×c+c×a=20Area of Δ=12AB×ACSince AB=b-a and AC=c-ai.e. AB ×AC=b-a×c-a                   =b×c-b×a-a×c+a×a                   =b×c+a×b+c×a+0                            a×b=-b×a     a×c=-c×a     a×a=0                  =b×c+a×b+c×ai.e. 12AB ×AC=12a×b+b×c+c×a                        =12×20      giveni.e. Area of ΔABC=10

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Question 9:

If a , b are two vectors such that a=1,b=2 and a×b=2 then the angle between a and b is _____________.

Answer:

For given vectors a and ba=1, b=2 and a×b=2 By lagrange's identity, a×b2+a.b2=a2 b2i.e. 2+a.b2=1×4i.e. a.b2=2i.e. a.b=2 ab cosθ=2      where θ is angle between a and bi.e. 1×2 cosθ=2i.e. cosθ=12i.e. θ=π4 or 3π4 i.e angle between a and b is π4or 3π4

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Question 10:

For any two-collinear vectors a and b, the value of a.a×b is ____________.

Answer:

For noncollinear vector a and b.a.a×bLet a=a1i^+a2j^+a3k^      b=b1i^+b2j^+b3k^then a×b=i^j^k^a1a2a3b1b2b3a×b=i^a2 b3-a3 b2-j^a1b3-a3b1+k^a1 b2-a2 b1then a·a×b=a1i^+a2j^+a3k^·i^a2 b3-a3 b2-j^a1 b3-a3 b1+k^a1 b2-a2 b1=a1 a2 b3-a3 b2+a2a3 b1-a1 b3+a3a1 b2-a2 b1=a1a2 b3-a1a3 b2+a2a3 b1-a2a1 b3+a3a1 b2-a3a2 b1i.e. a.a×b=0

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Question 11:

If a and b are two non-zero non-collinear vectors such that a×b=1 and a . b=3, then the angle between a and b is _____________.

Answer:

For nonzero non collinear vectors a and bGiven a×b=1            a·b=3Since by lagrange's identity,a×b2+a.b2=a2 b2i.e. 1+3=a2 b2i.e. 4=a2 b2i.e. a b=2a.b=abcosθ where θ is angle between a and b3=2cosθi.e. cosθ=32i.e. θ=π6

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Question 12:

If three points with position vectors a, b and c are collinear, then a×b+b×c+c×a= _______________.

Answer:

If a, b and c are collinear,

Let points A, B, C be collinear, where position vectors are a, b and c respectively, since AB and BC are parallel vectors, 

i.e AB×BC=0i.e b-a×c-b=0i.e b×c+b×-b+-a×c+a×b=0i.e b×c+0+c×a+a×b=0i.e a×b+b×c+c×a=0

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Question 1:

Define vector product of two vectors.

Answer:

If a and b are two non-zero non-parallel vectors, then the vector product denoted by a×b is defined as a×b=a b sin θ η^.Here, θ is the angle between a and b and η^ is the unit vector perpendicular to the plane of a and b such that a, b and η^ form a right handed system.

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Question 2:

Write the value i^×j^·k^+i^·j^.

Answer:

i^×j^. k^+i^. j^=k^. k^ +0=k^2+0=12+0           ( k=1)=1

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Question 3:

Write the value of i^.j^×k^+j^.k^×i^+k^.j^×i^.

Answer:

i^. j^×k^+j^. k^×i^+k^. j^×i^=i^. i^+j^. j+k^. -k^=i^2+j^2-k^2 =1+1-1                   ( i^=1, j^=1 and k^=1)=1

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Question 4:

Write the value of i^.j^×k^+j^.k^×i^+k^.i^×j^.

Answer:

i^. j^×k^+j^. k^×i^+k^. i^×j^=i^. i^+j^. j+k^. k^=i^2+j^2+k^2=1+1+1=3

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Question 5:

Write the value of i^×j^+k^+j^×k^+i^+k^×i^+j^.

Answer:

i^× j^+k^+j^× k^+i^+k^× i^+j^=i^× j^+i^× k^+j^× k^+j^× i^+k^× i^+k^× j^=k^-j^+i^-k^+j^-i^=0

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Question 6:

Write the expression for the area of the parallelogram having a and b as its diagonals.

Answer:

Given: a and b are diagonals of a parallelogram.Area of the parallelogram = 12 a×b

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Question 7:

For any two vectors a and b write the value of  a .b 2+ a ×b 2 in terms of their magnitudes.

Answer:

a. b2+a×b2=a b cos θ2+a b sin θ2=a2 b2 cos2 θ+sin2 θ=a2 b2 1=a2 b2 

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Question 8:

If a and b are two vectors of magnitudes 3 and 23 respectively such that a ×b is a unit vector. Write the angle between a and b .

Answer:

Let θ be the angle between aand b.It is given that a×b is a unit vector.a×b=1We know a×b=a b sin θ1=3 23 sin θsin θ=12θ=45o, 135o

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Question 9:

If  a =10, b =2 and  a ×b =16, find a .b .

Answer:

We know a.b2+a ×b2=a2b2a.b2+162=102×22          ( a×b=16, a=10 and b=2)a.b2+256=400a.b2=144a.b=±12

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Question 10:

For any two vectors a and b , find a . b ×a .

Answer:

Let: a=a1i^+a2j^+a3k^b=b1i^+b2j^+b3k^b×a=i^j^k^b1b2b3a1a2a3         =i^ b2a3-b3a2-j^ b1a3-b3a1+k^ b1a2-b2a1Now,a. b×a=a1i^+a2j^+a3k^. i^ b2a3-b3a2-j^ b1a3-b3a1+k^ b1a2-b2a1=a1b2a3-b3a2-a2 b1a3-b3a1+a3 b1a2-b2a1=a1b2a3-a1b3a2-a2b1a3+a2b3a1+a3b1a2-a3b2a1=0

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Question 11:

If a and b are two vectors such that  a ×b =3and a .b =1, find the angle between.

Answer:

a×b=3a b sin θ=3        ...(1)a. b=1a b cos θ=1           ...(2)Dividing (1) by (2), we geta b sin θa b cos θ=3tan θ=3θ=60o



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Question 12:

For any three vectors a, b and c write the value of a × b +c +b × c +a +c × a +b .

Answer:

a×b+c+b×c+a+c×a+b=a×b+a×c+b×c+b×a+c×a+c×b=a×b+a×c+b×c-a×b-a×c-b×c=0

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Question 13:

For any two vectors a and b , find  a ×b  . b .

Answer:

Let:a=a1i^+a2j^+a3k^b=b1i^+b2j^+b3k^a×b=i^j^k^a1a2a3b1b2b3=i^ a2b3-a3b2-j^ a1b3-a3b1+k^ a1b2-a2b1a×b. b=i^ a2b3-a3b2-j^ a1b3-a3b1+k^ a1b2-a2b1 . b1i^+b2j^+b3k^=b1 a2b3-a3b2 -b2 a1b3-a3b1+b3 a1b2-a2b1=a2b1b3-a3b1b2-a1b2b3+a3b1b2+a1b2b3-a2b1b3=0

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Question 14:

Write the value of i^×j^×k^.

Answer:

i^×j^×k^=i^×i^=0

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Question 15:

If a =3i^-j^+2k^ and b =2i^+j^-k^, then find  a ×b  a .

Answer:

Since a×b is a vector, a×b a without any dot or cross product in between is meaningless.

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Question 16:

Write a unit vector perpendicular to i^+j^ and j^+k^.

Answer:

Let a=i^+j^+0k^; b=0i^+j^+k^a×b=i^j^k^110011         = i^-j^+k^a×b=1+1+1                =3Unit vector perpendicular to a and b is,a×ba×b=13i^-j^+k^

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Question 17:

If  a ×b 2+ a .b 2=144 and  a =4, find  b .

Answer:

We know a×b2+a. b2=a2 b2144=42 b2144=16 b2b2=9b=3

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Question 18:

If r =xi^+yj^+zk^, then write the value of r ×i^2.

Answer:

Given:r=xi^+yj^+zk^Now,i=i^+0j^+0k^r×i=i^j^k^xyz100         =0 i^+zj^-yk^r×i=z2+y2r×i2=z2+y2

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Question 19:

If a and b are unit vectors such that a ×b is also a unit vector, find the angle between a and b .

Answer:

Let θ be the angle between a and b.Given:a×b=1a=1b=1We knowa×b=a b sin θ1=1 1 sin θsin θ=1 θ=π2

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Question 20:

If a and b are two vectors such that  a .b = a ×b , write the angle between a and b .

Answer:

Let θ be the angle between a and b.We know a×b=a b sin θa. b=a b cos θNow,a×b=a. b            (Given)a b sin θ=a b cos θsin θ=cos θθ=π4

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Question 21:

If a and b are unit vectors, then write the value of  a ×b 2+ a .b 2.

Answer:

It is given that a and b are unit vectors.a=b=1         ...(1)Now,a. b2+a×b2=a b cos θ2+a b sin θ2=a2 b2 cos2 θ+sin2 θ=a2 b2 1=a2 b2 =12 12                 [From (1)]=1

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Question 22:

If a is a unit vector such that a ×i^=j^, find a .i^.

Answer:

We know k^×i^=j^                      ...(1)Given: a×i^=j^         ...(2)               Comparing (1) and (2), we geta=k^Now,a. i^=k^. i^      =0

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Question 23:

If c is a unit vector perpendicular to the vectors a and b ,write another unit vector perpendicular to a and b .

Answer:

c is a unit vector perpendicular to both a and b.c=a×ba×b⇒-c=b×aa×bTherefore, -c is perpendicular to b and a.Thus,-c is another unit vector perpendicular to a and b.

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Question 24:

Find the angle between two vectors a and b with magnitudes 1 and 2 respectively and when  a ×b =3.

Answer:

Let θ be the angle between a and b.We know a×b=a b sin θ3=1 2 sin θsin θ=32θ=π3

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Question 25:

Vectors a and b are such that  a =3, b =23and a ×b  is a unit vector. Write the angle between a and b .

Answer:

Given: a×b is a unit vector.a×b=1              ...(1)Let θ be the angle between aand b.We know a×b=a b sin θFrom (1), we get1=3 23 sin θ    sin θ=32θ=π3

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Question 26:

Find λ, if 2i^+6j^+14k^×i^-λj^+7k^=0 .

Answer:

Given: i^j^k^26141-λ7=0i^ 42+14λ-0j^+k^ -2λ-6=0i^+0j^+0k^42+14λ=0; -2λ-6=0λ=-3      (This satisfies the above equations) 

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Question 27:

Write the value of the area of the parallelogram determined by the vectors 2i^ and 3j^.

Answer:

Let:a=2i^b=3j^a×b=6 i^×j^          =6k^Area of the parallelogram=a×b=6 k^=61=6 sq. units

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Question 28:

Write the value of i^×j^·k^+j^+k^·j^

Answer:

i^×j^. k^+j^+k^. j^=k^. k^ +j^. j^+k^. j^     ( i^×j^= k^) =k^2+j^2+0           ( k^. j^=0)=12+12=2

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Question 29:

Find a vector of magnitude 171 which is perpendicular to both of the vectors a=i^+2j^-3k^ and b=3i^-j^+2k^.

Answer:


The given vectors are a=i^+2j^-3k^ and b=3i^-j^+2k^.

Unit vectors perpendicular to both a and b = ±a×ba×b

Now,
a×b=i^j^k^12-33-12=i^-11j^-7k^a×b=i^-11j^-7k^=12+-112+-72=1+121+49=171

Unit vectors perpendicular to both a and b = ±i^-11j^-7k^171

∴ Required vectors = 171±i^-11j^-7k^171=±i^-11j^-7k^

Thus, the vectors of magnitude 171 which are perpendicular to both the given vectors are ±i^-11j^-7k^.

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Question 30:

Write the number of vectors of unit length perpendicular to both the vectors a=2i^+j^+2k^ and b=j^+k^.

Answer:

Unit vectors perpendicular to a and b are ±a×ba×b.
a×b=i^j^k^212011=-i^-2j^+2k^ 

∴ Unit vectors perpendicular to a and b are ±-i^-2j^+2k^-12+-22+22=±-13i^-23j^+23k^

Thus, there are two unit vectors perpendicular to the given vectors.

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Question 31:

Write the angle between the vectors a×b and b×a.

Answer:


b×a = -a×b 

So, a×b and b×a are vectors of same magnitude but opposite in directions. 

Thus, the angle between the vectors a×b and b×a is 180º.



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