Mathematics Multiple Choice Questions on “Product of Two Vectors-1”.
1. Find the scalar product of the vectors (vec{a}=2hat{i}+5hat{j}) and (vec{b}=6hat{i}-7hat{j}).
a) -32
b) -23
c) 32
d) 23
Answer: b
Clarification: If (vec{a} ,and ,vec{b}) are two vectors, where a1, a2 are the components of vector (vec{a} ,and ,b_1, ,b_2) are the components of vector (vec{b}), then the scalar product is given by
(vec{a}.vec{b}=a_1 ,b_1+a_1 ,b_2)
∴((2hat{i}+5hat{j}).(6hat{i}-7hat{j}))=2(6)+5(-7)=12-35=-23.
2. Find the angle between the two vectors (vec{a}) and (vec{b}) with magnitude (sqrt{3}) and (sqrt{2}) respectively and (vec{a.} ,vec{b}=3sqrt{2}).
a) (cos^{-1}frac{1}{sqrt{3}})
b) (cos^{-1}sqrt{3})
c) (cos^{-1}frac{3}{sqrt{2}})
d) (cos^{-1}frac{2}{sqrt{3}})
Answer: a
Clarification: Given that, (|vec{a}|=sqrt{3} ,and ,|vec{b}|=sqrt{2})
Also, (vec{a.} vec{b}=3sqrt{2})
The angle between two vectors is given by
(cosθ=frac{|vec{a}|.|vec{b}|}{vec{a}.vec{b}})
∴(cosθ=frac{sqrt{3}.sqrt{2}}{3sqrt{2}}=frac{1}{sqrt{3}})
∴(θ=cos^{-1}frac{1}{sqrt{3}}).
3. Find the projection of vector (vec{a}=8hat{i}-hat{j}+6hat{k}) on vector (vec{b}= 4hat{i}+3hat{j}).
a) (sqrt{frac{29}{5}})
b) (frac{29}{sqrt{5}})
c) (frac{sqrt{29}}{5})
d) (frac{29}{5})
Answer: d
Clarification: The projection of a vector (vec{a}) on vector (vec{b}) is given by
(frac{1}{|vec{b}|} (vec{a}.vec{b}))
(|vec{b}|=sqrt{4^2+3^2}=sqrt{16+9})=5
(vec{a}.vec{b})=8(4)-1(3)+0=32-3=29
The projection of vector (8hat{i}-hat{j}+6hat{k}) on vector (4hat{i}+3hat{j}) will be
(frac{1}{|vec{b}|} (vec{a}.vec{b})=frac{1}{5} (29)=frac{29}{5})
4. Find (|vec{a}+vec{b}|), if (|vec{a}|=3 ,and ,|vec{b}|=4 ,and ,vec{a}.vec{b}=6).
a) 34
b) (sqrt{37})
c) 13
d) (sqrt{23})
Answer: b
Clarification: (|vec{a}+vec{b}|^2=(vec{a}+vec{b}).(vec{a}+vec{b}))
=(vec{a}.vec{a}+vec{a}.vec{b}+vec{b}.vec{a}+vec{b}.vec{b})
=(|vec{a}|^2+2(vec{a}.vec{b})+|vec{b}|^2)
=(3)2+2(6)+(4)2
=9+12+16=37
∴(|vec{a}+vec{b}|=sqrt{37})
5. Find the angle between the vectors (vec{a}=hat{i}-hat{j}+2hat{k} ,and ,vec{b}=3hat{i}+2hat{j}+4hat{k}).
a) (cos^{-1}sqrt{frac{58}{3}})
b) (cos^{-1}frac{sqrt{58}}{3})
c) (cos^{-1}frac{58}{3sqrt{3}})
d) (cos^{-1}frac{sqrt{58}}{3sqrt{3}})
Answer: d
Clarification: The angle between the two vectors is given by
(cosθ=frac{|vec{a}|.|vec{b}|}{vec{a}.vec{b}})
(|vec{a}|=sqrt{1^2+(-1)^2+2^2}=sqrt{1+1+4}=sqrt{6})
(|vec{b}|=sqrt{3^2+2^2+(-4)^2}=sqrt{9+4+16}=sqrt{29})
(vec{a}.vec{b})=1(3)-1(2)+2(4)=9
∴(cosθ=frac{sqrt{6}.sqrt{29}}{9}=frac{sqrt{58}}{3sqrt{3}})
∴(θ=cos^{-1}frac{sqrt{58}}{3sqrt{3}})
6. Find the angle between the vectors (vec{a}=-hat{i}+hat{j}-hat{k}) and (vec{b}=hat{i}-hat{j})
a) (cos^{-1}-frac{sqrt{3}}{2})
b) (cos^{-1}-frac{2}{sqrt{3}})
c) (cos^{-1}-sqrt{2})
d) (cos^{-1}-sqrt{frac{3}{2}})
Answer: d
Clarification: The angle between two vectors is given by
(cosθ=frac{|vec{a}|.|vec{b}|}{vec{a}.vec{b}})
(|vec{a}|=sqrt{(-1)^2+(1)^2+(-1)^2}=sqrt{3})
(|vec{b}|=sqrt{(1)^2+(-1)^2}=sqrt{2})
(vec{a}.vec{b})=(-1)(1)+1(-1)+0=-2
(cosθ=frac{sqrt{3}.sqrt{2}}{-2}=-sqrt{frac{3}{2}})
∴(θ=cos^{-1}-sqrt{frac{3}{2}})
7. If two non-zero vectors (vec{a} ,and , vec{b}) are perpendicular to each other then their scalar product is zero.
a) True
b) False
Answer: a
Clarification: The given statement is true. If the angle between two vectors is (frac{π}{2}) i.e. they are perpendicular to each other, their scalar product will be zero.
8. Find the angle between the two vectors (vec{a} ,and , vec{b}) with magnitude 2 and (sqrt{3}) respectively and (vec{a.} , vec{b})=4.
a) (frac{π}{3})
b) (frac{π}{6})
c) (cos^{-1}frac{sqrt{2}}{3})
d) (cos^{-1}frac{2}{sqrt{3}})
Answer: b
Clarification: Given that, (|vec{a}|=2 ,and ,|vec{b}|=sqrt{3})
Also, (vec{a.} ,vec{b}=4)
The angle between two vectors is given by
(cosθ=frac{|vec{a}|.|vec{b}|}{vec{a}.vec{b}})
∴(cosθ=frac{2.sqrt{3}}{4}=frac{sqrt{3}}{2})
∴(θ=cos^{-1}frac{sqrt{3}}{2}=frac{π}{6}).
9. Find the scalar product of the vectors (vec{a}=6hat{i}-7hat{j}+5hat{k} ,and ,vec{b}=6hat{i}-7hat{k})
a) 1
b) 8
c) 6
d) 3
Answer: a
Clarification: If (vec{a} ,and ,vec{b}) are two vectors, where (a_1, a_2, a_3) are the components of vector (vec{a} ,and ,b_1, b_2, b_3) are the components of vector (vec{b}), then the scalar product is given by
(vec{a}.vec{b}=a_1 b_1+a_1 b_2+a_3 b_3)
((6hat{i}-7hat{j}+5hat{k}).(6hat{i}-7hat{k}))=6(6)-7(0)+5(-7)=36-35=1.
10. Find the projection of vector (vec{b}=2hat{i}+2sqrt{2} ,hat{j}-2hat{k}) on the vector (vec{a}=hat{i}-hat{j}-sqrt{2} ,hat{k}).
a) 2
b) (sqrt{2})
c) 1
d) (2sqrt{2})
Answer: b
Clarification: The projection of vector (vec{b}) on the vector (vec{b}) is given by (frac{1}{|vec{a}|} (vec{a}.vec{b}))
(|vec{a}|=sqrt{(1)^2+(-1)^2+(-sqrt{2})^2}=sqrt{1+1+2}=sqrt{4})=2
Also, (vec{a}.vec{b}=2(1)+2sqrt{2} ,(-1)-2(-sqrt{2})=2-2sqrt{2}+2sqrt{2})=2
Therefore, the projection of vector (hat{i}-hat{j}-sqrt{2} ,hat{k}) on the vector (vec{b}=2hat{i}+2sqrt{2}hat{j}-2hat{k}) is
(frac{1}{|vec{a}|} (vec{a}.vec{b})=frac{1}{2}) (2)=1.