Monday, August 23, 2021

Vector Algebra: Mathematics MCQ

 

Q.1. The vector  straight a with rightwards arrow on top equals straight alpha straight i with hat on top plus 2 straight j with hat on top plus straight beta straight k with hat on top lies in the plane of the vectors straight b with rightwards arrow on top equals straight i with hat on top plus straight j with hat on top space and space straight c with rightwards arrow on top equals straight j with hat on top plus straight k with hat on top and bisect the angle between b with rightwards arrow on top space a n d space c with rightwards arrow on top . Then which one of the following gives possible values of straight alpha space and space straight beta    (AIEEE 2008) 

A) alpha equals 1 comma space beta equals 2

B) alpha equals 2 comma space beta equals 1

C) alpha equals 1 comma space beta equals 1

D) alpha equals 2 comma space beta equals 2 


Q2. If 4i+7j+8k, 2i+3j+4k and 2i+5j+7k are the position vectors of the vertices A,B and C respectively of triangle ABC. The position vectors of the point where the bisectors of angle A meets BC is... (Pb. CET 2004)

A) 1 third left parenthesis 6 i plus 13 j plus 18 k right parenthesis

B) 2 over 3 left parenthesis 6 i _ 12 j minus 8 k right parenthesis

C) 1 third left parenthesis negative 6 i minus 8 j minus 9 k right parenthesis

D) 2 over 3 left parenthesis negative 6 i plus 12 j plus 8 k right parenthesis 


Q.3. Direction of a reciprocal vector of a vector a with rightwards arrow on top is    ( MP PET 2013)

a) Same as that of a with rightwards arrow on top

B) Opposite to a with rightwards arrow on top

C) Perpendicular to a with rightwards arrow on top

D) None of these


Q.4. If top enclose a equals i with hat on top plus j with hat on top plus k with hat on top comma space top enclose b equals 2 i with hat on top plus lambda j with hat on top plus k with hat on top comma space top enclose c equals i with hat on top minus j with hat on top plus 4 k with hat on top space space a n d space top enclose a. left parenthesis top enclose b cross times top enclose c right parenthesis equals 10 , then 

A) 6

B) 7

C) 9

D) 10

Explanation-

open square brackets a with bar on top space b with bar on top space c with bar on top close square brackets equals 10
open vertical bar
1 space space space space 1 space space space space space space 1
2 space space space space lambda space space space space space space 1
1 space space space minus 1 space space space space 4 space close vertical bar equals 10
left parenthesis 4 lambda plus 1 right parenthesis minus left parenthesis 8 minus 1 right parenthesis plus left parenthesis negative 2 minus lambda right parenthesis equals 10
4 lambda plus 1 minus 7 minus 2 minus lambda equals 10
3 lambda equals 10 plus 8 equals 18
lambda equals 6.             


Q.5. If the position vector of three points are straight a minus 2 straight b plus 3 straight c comma space 2 straight a plus 3 straight b minus 4 straight c comma negative 7 straight b plus 10 straight c comma then the three points are 

A) Collinear 

B) Non-coplanar

C) Non-collinear

D) Non of these 

Explanation-

Let P,Q, R be the given points. Then, we have 

                   QR=-2PQ

Hence, P,Q and R are collinear.


Q.6. If the vectors negative straight i with hat on top plus 3 straight j with hat on top minus 4 straight k with hat on top space and space 2 straight i with hat on top minus 6 straight j with hat on top plus straight lambda straight k with hat on top are collinear then find the value of straight lambda 

A) 4

B) 6

C) 8

D) 10

Explanation-

Let space top enclose straight alpha equals negative straight i with hat on top plus 3 straight j with hat on top minus 4 straight k with hat on top space and space top enclose straight b equals 2 straight i with hat on top minus 6 straight j with hat on top plus straight lambda straight k with hat on top
As space top enclose straight alpha space and space top enclose straight b space are space collinear space comma space top enclose straight alpha cross times straight b with minus on top equals 0
therefore space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row cell negative 1 end cell 3 cell negative 4 end cell row 2 cell negative 6 end cell straight lambda end table close vertical bar equals 0
therefore space straight i with hat on top open parentheses 3 straight lambda minus 24 close parentheses minus straight j with hat on top open parentheses negative straight lambda plus 8 close parentheses plus straight k with hat on top open parentheses 4 minus 6 close parentheses equals top enclose 0
rightwards double arrow space 3 open parentheses straight lambda minus 8 close parentheses straight i with hat on top plus open parentheses straight lambda minus 8 close parentheses straight j with hat on top plus 0 straight k with hat on top equals 0 straight i with hat on top plus 0 straight j with hat on top plus 0 straight k with hat on top
rightwards double arrow space straight lambda minus 8 equals 0 space space space space space space space space space space space space rightwards double arrow straight lambda equals 8 


Q.7. If the origin is the centroid of the triangle whose vertices are A (a, 2, -1) , B (5, b, 2) and C (-3, 1, (C) then find the values of a , b and c .

A) a = 2 , b = 3 , c = 1

B) a = 1 , b = 2 , c = 3

C) a = - 2 , b = - 3 , c = 1

D) a = 2 , b = 3 , c = 1

Explanation-

Let space top enclose straight alpha equals straight alpha straight i with hat on top plus 2 straight j with hat on top minus straight k with hat on top comma space space space space space top enclose straight b equals 5 straight i with hat on top plus straight b straight j with hat on top plus 2 straight k with hat on top space space space and space space top enclose straight c equals negative 3 straight i with hat on top plus straight j with hat on top plus straight c straight k with hat on top.
If space straight O open parentheses top enclose straight o close parentheses space is space the space centriod space of space the space triangle space then space by space centriod space formula space.
top enclose straight o equals fraction numerator top enclose straight alpha plus top enclose straight b plus top enclose straight c over denominator 3 end fraction
space space equals fraction numerator open parentheses straight alpha plus 2 close parentheses straight i with hat on top plus left parenthesis 3 plus straight b right parenthesis straight j with hat on top plus open parentheses 1 plus straight c close parentheses straight k with hat on top over denominator 3 end fraction
space space equals fraction numerator open parentheses straight alpha straight i with hat on top plus 2 straight j with hat on top minus straight k with hat on top close parentheses plus open parentheses 5 straight i with hat on top plus straight b straight j with hat on top plus 2 straight k with hat on top close parentheses plus open parentheses negative 3 straight i with hat on top plus straight j with hat on top plus straight c straight k with hat on top close parentheses over denominator 3 end fraction
top enclose straight o space equals open parentheses fraction numerator straight alpha plus 2 over denominator 3 end fraction close parentheses straight i with hat on top plus open parentheses fraction numerator 3 plus straight b over denominator 3 end fraction close parentheses straight j with hat on top plus open parentheses fraction numerator 1 plus straight c over denominator 3 end fraction close parentheses straight k with hat on top
rightwards double arrow space fraction numerator straight alpha plus 2 over denominator 3 end fraction equals 0 space comma space fraction numerator 3 plus straight b over denominator 3 end fraction equals 0 comma space fraction numerator 1 plus straight c over denominator 3 end fraction equals 0
rightwards double arrow space straight alpha equals negative 2 comma space straight b equals negative 3 comma space straight c equals negative 1 space 


Q.8. If O be the origin and the position vector of A be 4 i +5 j , then a unit vector parallel to stack O A with rightwards arrow on top is

A) fraction numerator 3 over denominator square root of 41 end fraction open parentheses straight i plus 4 straight j close parentheses

B) fraction numerator 4 over denominator square root of 41 end fraction open parentheses straight i minus 6 straight j close parentheses

C) fraction numerator 1 over denominator square root of 41 end fraction open parentheses 4 space straight i space plus 5 space straight j close parentheses

D) fraction numerator 1 over denominator square root of 31 end fraction open parentheses 4 space straight i space minus 5 space straight j close parentheses 

Explanation-

OA with hat on top equals fraction numerator 4 straight i plus 5 straight j over denominator square root of 16 plus 25 end root end fraction equals fraction numerator 1 over denominator square root of 41 end fraction open parentheses 4 straight i plus 5 straight j close parentheses 


Q.9. If top enclose straight a space equals space 5 straight i with hat on top space minus space 3 straight j with hat on top space plus space 4 straight k with hat on top space comma space top enclose straight b space equals space space straight i with hat on top space minus space 2 straight j with hat on top space plus space straight k with hat on top comma space top enclose straight c space equals space 3 straight i with hat on top space plus space 5 straight j with hat on top then find top enclose straight a space. space left parenthesis top enclose straight b space cross times space top enclose straight c right parenthesis 

A) 8

B) 10

C) 14

D) 20

Explanation-

We have, top enclose straight a space equals space 5 straight i with hat on top space minus space 3 straight j with hat on top space plus space 4 straight k with hat on top space comma space top enclose straight b space equals space straight i with hat on top space minus space 2 straight j with hat on top space plus space straight k with hat on top space comma space top enclose straight c space equals space 3 straight i with hat on top space plus space 5 straight j with hat on top space

therefore space top enclose straight a space. space left parenthesis top enclose straight b space cross times space top enclose straight c space right parenthesis space equals space left square bracket top enclose straight a space top enclose straight b space top enclose straight c right square bracket
equals space open vertical bar table row 5 cell negative 3 end cell 4 row 1 cell negative 2 end cell 1 row 3 5 0 end table close vertical bar
equals space 5 space left parenthesis 0 minus 5 right parenthesis space plus space 3 space left parenthesis 0 minus 3 right parenthesis space plus 4 space left parenthesis 5 space plus 6 right parenthesis
equals space minus 25 minus space 9 space plus space 44 space equals 10 


Q.10. Find the vector of magnitude 3 units which is parallel to 4 straight i with hat on top plus 3 straight j with hat on top .

A) 2/5 (4 straight i with hat on top plus 3 straight j with hat on top)

B) 3/5(4 straight i with hat on top plus 3 straight j with hat on top)

C) 7/5(4 straight i with hat on top plus 3 straight j with hat on top)

D) 3/5(4 straight i with hat on top plus 3 straight j with hat on top)

Explanation-

top enclose straight a equals 4 straight i with hat on top plus 3 straight j with hat on top equals open vertical bar top enclose straight a close vertical bar equals square root of 4 squared plus 3 squared end root equals square root of 16 plus 9 end root equals square root of 25 equals 5
We comma space have comma space straight a with hat on top equals fraction numerator top enclose straight a over denominator open vertical bar top enclose straight a close vertical bar end fraction equals fraction numerator 4 straight i with hat on top plus 3 straight j with hat on top over denominator 5 end fraction
therefore The space required space vector space is space 3 top enclose straight a equals 3 divided by 5 left parenthesis 4 straight i with hat on top plus 3 straight j with hat on top right parenthesis  


                              


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