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.
Exercise 10.3
Q1. Find the angle between two vectors →a and →b with magnitudes √3 and 2, respectively having →a⋅→b=√6
Answer. |→a|=√3,|→b|=2 and, →a⋅→b=√6 Now, we know that ¯¯¯a⋅→b=|→a||¯¯b|cosθ∴√6=√3×2×cosθ⇒cosθ=√6√3×2⇒cosθ=1√2⇒θ=π4 Hence , the angle between the given vectors →a and →b is π4
Q2. Find the angle between the given vector ^i−2^j+3^k and 3^i−2^j+^k
Answer. The qiven vectors are ^a=^i−2^j+3^k and →b=3^i−2^j+^k|→a|=√12+(−2)2+32=√1+4+9=√14|→b|=√32+(−2)2+12=√9+4+1=√14 Now, →a⋅→b=(^i−2^j+3^k)(3^i−2^j+^k)=3+4+3=10
Q3. Find the projection of the vector ^i−^j on the vector ^i+^j
Answer. Let ¯¯¯a=^i−^j and →b=^i+^j Now, projection of vector →a on →b is given by 1|→b|(→a→b)=1√1+1{1.1+(−1)(1)}=1√2(1−1)=0 Hence the projection of vector →a on →b is 0.
Q4. Find the projection of the vector ^i+3^j+7^k on the vector 7^i−^ȷ+8^k
Answer. Let ¯¯¯a=^i+3^j+7^k and ^b=7^i−^j+8^k Now, projection of vector →a on →b is given by Now, projection of vector →aon→b is given by 1|→b|(→a⋅→b)=1√72+(−1)2+82{1(7)+3(−1)+7(8)}=7−3+56√49+1+64=60√114
Q5. Show that each of the given three vectors is a unit vector: 17(2^i+3^j+6^k)⋅17(3^i−6^j+2^k),17(6^i+2^j−3^k). Also, show that they are mutually perpendicular to each other.
Answer. Let →a=17(2^i+3^j+6^k)=27^i+37^j+67^k→b=17(3^i−6^j+2^k)=37^i−67^j+27^k¯¯c=17(6^i+2^j−3^k)=67^i+27^j−37^k |→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. →a⋅→b=27×37+37×(−67)+67×27=649−1849+1249=0→b⋅→c=37×67+(−67)×27+27×(−37)=1849−1249−649=0→c⋅→a=67×27+27×37+(−37)×67=1249+649−1849=0 Hence, the given three vectors are mutually perpendicular to each other.
Q6. Find |→a| and |→b|, if (→a+→b)⋅(→a−→b)=8 and |→a|=8|→b|
Answer. (→a⋅→b)⋅(→a−→b)=8⇒→a⋅→a−→a→b+→b⋅→a−→b⋅→b=8⇒|→a|2−|→b|2=8⇒(8|→b|)2−|→b|2=8⇒64|→b|2−|→b|2=8⇒63|→b|2=8 ⇒|→b|2=863⇒|→b|=√863⇒|→b|=2√23√7|→a|=8|→b|=8×2√23√7=16√23√7
Q7. Evaluate the product (3→a−5→b)⋅(2→a+7→b)
Answer. (3→a−5→b)⋅(2→a+7→b)=3→a⋅2→a+3→a⋅7→b−5→b⋅2→a−5→b⋅7→b=6→a⋅→a+21→a⋅→b−10→a⋅→b−35→b⋅→b=6|→a|2+|1→a⋅→b−35|→b∣∣2
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|,→a⋅→b=12, and θ=60∘ . We know that ¯¯¯a⋅→b=|→a||→b|cosθ we know that ¯¯¯a⋅→b=|→a||→b|cosθ∴12=|→a||→a|cos60∘⇒12=|→a|2×12⇒1|a|2=1⇒|→a|=|→b|=1
Q9. Find|→x|, if for a unit vector →a,(→x−→a)⋅(→x+→a)=12
Answer. (→x−→a)⋅(→x+→a)=12⇒→x⋅→x+→x⋅→a−→a⋅→x−¯¯¯a⋅→a=12⇒|→x|2−|→a|2=12⇒|→x|2−1=12[|→a|=1 as →a is a unit vector ]⇒|→x|2=13∴|→x|=√13
Q10. If →a=2^i+2^j+3^k,→b=−^i+2^j+^k and →c=3^i+^j are such that →a+λ→b is perpendicular to →c , then find the value of λ .
Answer. The given vectors are ¯¯¯a=2^i+2^j+3^k,→b=−^i+2^j+^k, and →c=3^i+^j Now, →a+λ→b=(2^i+2^j+3^k)+λ(−^i+2^j+^k)=(2−λ)^i+(2+2λ)^j+(3+λ)^k If (→a+λ→b) is perpendicular to ¯¯c, then (→a+λ→b)⋅→c=0 ⇒[(2−λ)^i+(2+2λ)^j+(3+λ)^k]⋅(3^i+^j)=0⇒(2−λ)3+(2+2λ)1+(3+λ)0=0⇒6−3λ+2+2λ=0⇒−λ+8=0⇒λ=8 Hence, the required value of λ is 8 .
Q11. Show that |→a|→b+|→b|→a is perpendicular to 0|→a|→b−|→b|→a, for any two non-zero vectors →a and →b
Answer. (|→a|→b+|→b|→a)⋅(|→a|→b−|→b|→a)=|→a|2→b⋅→b−|→a||→b|→b⋅→a+|→b||→a⋅→b−|→b∣∣2→a⋅→a=|→a|2|→b|2−|→b|2|→a|2=0
Q12. If¯¯¯a⋅→a=0 and →a⋅→b=0, then what can be concluded about the vector →b
Answer. It is given that →a⋅→a=0 and →a⋅→b=0 Now, →a⋅→a=0⇒|→a|2=0⇒|→a|=0∴→a is a zero vector. Hence, vector →b satisfying →a⋅→b=0 can be any vector.
Q13. If →a,→b,→c are unit vectors such that →a+→b+→c=¯¯¯0, find the value of →a⋅→b+→b⋅→c+→c⋅→c⋅→a
Answer. (|→a|→b+|→b|→a)⋅(|→a|→b−|→b|→a)=|→a|2→b⋅→b−|→a||→b|→b⋅→a+|→b||→a⋅→b−|→b∣∣2→a⋅→a=|→a|2|→b|2−|→b|2|→a|2=0
Q14. If either vector →a=→0 or →b=→0 , then →a⋅→b=0 . But the converse need not be true. Justify your answer with an example.
Answer. Consider →a=2^i+4^j+3^k and →b=3^i+3^j−6^k Then, →a⋅→b=2.3+4.3+3(−6)=6+12−18=0 we now observe that: |→a|=√22+42+32=√29∴→a=√32+32+(−6)2=√54∴→b=→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+(2−0)^j+(3−0)^k=2^i+2^j+3^k¯¯¯¯¯¯¯¯BC={0−(−1)}^i+(1−0)^j+(2−0)^k=^i+^j+2^k∴−−→BA⋅−−→BC=(2^i+2^j+3^k)⋅(^i+^j+2˙k)=2×1+2×1+3×2=2+2+6=10 Now, it is known that: −−→BA⋅−−→BC=|BA∥BC|cos(∠ABC)∴10=√17×√6cos(∠ABC)⇒cos(∠ABC)=10√17×√6⇒∠ABC=cos−1(10√102)
Q16. Show that the points A (1, 2, 7), 3 (2, 6, 3) and C (3, 10, —1) are collinear.
Answer. |−−→AB=(2−1)^i+(6−2)^j+(3−7)^k=^i+4^j−4^k∴−−→AC=(3−2)^i+(10−6)^j+(−1−3)^k=^i+4^j−4^k−−→AC=(3−1)^i+(10−2)^j+(−1−3)^k=2^i+8^j−8^k|−−→AB|=√12+42+(−4)2=√1+16+16=√33|−−→AC|=√12+42+82=√4+16+16=√33∴|−−→AC|=|−−→AB|+|−−→BC| ∴|−−→AC|=|−−→AB|+|−−→BC| Hence, the given points A,B , and C are collinear.
Q17. Show that the vector 2^i−^j+^k,^i−3^j−5^k and 3^i−4^j−4^k from the vertical of a right angled triangle.
Answer. Let vectors 2^i−^j+^k,^i−3^j−5^k and 3^i−4^j−4^k be position vector of point A,B,C respectively. ∴−−→AB=(1−2)^i+(−3+1)^j+(−5−1)^k=−^i−2^j−6^k¯¯¯¯¯¯¯¯BC=(3−1)^i+(−4+3)^j+(−4+5)^k=2^i−^j+^k¯¯¯¯¯¯¯¯AC=(2−3)^i+(−1+4)^j+(1+4)^k=−^i+3^j+5^k |−−→AB|=√(−1)2+(−2)2+(−6)2=√1+4+36=√41|BC|=√22+(−1)2+l2=√4+1+1=√6|−−→AC|=√(−1)2+32+52=√1+9+25=√35∴|BC|2+|−−→AC|2=6+35=41=|AB|2
Q18. →a is a nonzero vector of magnitude 'a' and λ a nonzero scalar, then λ→a is unit vector if (A)=1(B) λ=−1(C) a=|λ|(D) a=1|λ|(A)=1(B) λ=−1(C) a=|λ|(D) a=1|λ|
Answer. Vector λ→a is a unit vector if Now, |λ→a|=1⇒|λ||→a|=1⇒|→a|=1|λ|⇒a=1|λ| a=1|λ| Hence, vector λ→a is a unit vector if a=1|λ|
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 Exercise 10.3){kind=link}