NCERT Solutions Class 12 Maths Chapter-10 (Vector Algebra) Exercise 10.3

NCERT Solutions Class 12 Maths Chapter-10 (Vector Algebra) Exercise 10.3

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-10 (Vector Algebra)Exercise 10.3 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Solutions Class 12 Maths Chapter-10 (Vector Algebra) Exercise 10.3

Exercise 10.3

Q1. 

Answer. |a|=3,|b|=2 and, ab=6 Now, we know that a¯b=|a||b¯|cosθ6=3×2×cosθcosθ=63×2cosθ=12θ=π4 Hence , the angle between the given vectors 

Q2. Find the angle between the given vector 

Answer. 

NCERT Solutions Class 12 Maths Chapter-10 (Vector Algebra) Exercise 10.3

Q3. Find the projection of the vector 

Answer.  Let a¯=i^j^ and b=i^+j^ Now, projection of vector a on b is given by 1|b|(ab)=11+1{1.1+(1)(1)}=12(11)=0 Hence the projection of vector 

Q4. Find the projection of the vector 

Answer.  Let a¯=i^+3j^+7k^ and b^=7i^j^+8k^ Now, projection of vector a on b is given by  

Q5. Show that each of the given three vectors is a unit vector: 17(2i^+3j^+6k^)17(3i^6j^+2k^),17(6i^+2j^3k^). Also, show that they are mutually perpendicular to each other.

Answer.  Let a=17(2i^+3j^+6k^)=27i^+37j^+67k^b=17(3i^6j^+2k^)=37i^67j^+27k^c¯=17(6i^+2j^3k^)=67i^+27j^37k^ |a|=(27)2+(37)2+(67)2=449+949+3649=1|b|=(37)2+(67)2+(27)2=949+3649+949=1|c|=(67)2+(27)2+(37)2=3649+449+949=1 Thus, each of the given three vectors is a unit vector.  

Q6. Find 

Answer. (ab)(ab)=8aaab+babb=8|a|2|b|2=8(8|b|)2|b|2=864|b|2|b|2=863|b|2=8 

Q7. Evaluate the product 

Answer. 

Q8. Find the magnitude of two vectors a and b , having the same magnitude and such that the angle between them is 60 and their scalar product is 12.

Answer.  Let θ be the angle between the vectors a and b . |a|=|b|,ab=12, and θ=60 .  We know that a¯b=|a||b|cosθ 

Q9. 

Answer. 

Q10.  If a=2i^+2j^+3k^,b=i^+2j^+k^ and c=3i^+j^ are such that a+λb is perpendicular to c ,  then find the value of λ . 

Answer.  The given vectors are a¯=2i^+2j^+3k^,b=i^+2j^+k^, and c=3i^+j^ Now, a+λb=(2i^+2j^+3k^)+λ(i^+2j^+k^)=(2λ)i^+(2+2λ)j^+(3+λ)k^ If (a+λb) is perpendicular to c¯, then (a+λb)c=0 

Q11. Show that |a|b+|b|a is perpendicular to 0|a|b|b|a, for any two non-zero vectors 

Answer. 

Q12. 

Answer. 

Q13. 

Answer. 

Q14.  If either vector a=0 or b=0 , then ab=0 . But the converse need not be true. Justify  your answer with an example. 


Answer.  Consider a=2i^+4j^+3k^ and b=3i^+3j^6k^ Then, ab=2.3+4.3+3(6)=6+1218=0 we now observe that: |a|=22+42+32=29a=32+32+(6)2=54b=3 Hence, the converse of the given statement need not be true.

Q15. If the vertices A, B, C of a triangle ABC are (1, 2, 3), (—1, 0, 0), (0, 1, 2), respectively, then find DABC. [ABC is the angle between the vectors BA and BC ]

Answer. BA={1(1)}i^+(20)j^+(30)k^=2i^+2j^+3k^BC¯={0(1)}i^+(10)j^+(20)k^=i^+j^+2k^BABC=(2i^+2j^+3k^)(i^+j^+2k˙)=2×1+2×1+3×2=2+2+6=10 

Q16. Show that the points A (1, 2, 7), 3 (2, 6, 3) and C (3, 10, —1) are collinear.

Answer. |AB=(21)i^+(62)j^+(37)k^=i^+4j^4k^AC=(32)i^+(106)j^+(13)k^=i^+4j^4k^AC=(31)i^+(102)j^+(13)k^=2i^+8j^8k^|AB|=12+42+(4)2=1+16+16=33|AC|=12+42+82=4+16+16=33|AC|=|AB|+|BC| 

Q17. Show that the vector 2i^j^+k^,i^3j^5k^ and 3i^4j^4k^ from the vertical of a right angled triangle.

Answer. Let vectors 2i^j^+k^,i^3j^5k^ and 3i^4j^4k^ be position vector of point A,B,C respectively. AB=(12)i^+(3+1)j^+(51)k^=i^2j^6k^BC¯=(31)i^+(4+3)j^+(4+5)k^=2i^j^+k^AC¯=(23)i^+(1+4)j^+(1+4)k^=i^+3j^+5k^ 

Q18. a is a nonzero vector of magnitude 'a' and λ a nonzero scalar, then λa is unit vector if  (A)=1(B) λ=1(C) a=|λ|(D) a=1|λ|


Answer.  Vector λa is a unit vector if  Now, |λa|=1|λ||a|=1|a|=1|λ|a=1|λ| 

Chapter-10 (Vector Algebra)