Mathematics Question Papers for Class 12 on “Product of Two Vectors-2”.
1. Evaluate the product ((2vec{a}+5vec{b}).(4vec{a}-5vec{b})).
a) (|vec{a}|^2+2vec{a}.vec{b}-15|vec{b}|^2)
b) (8|vec{a}|^2+2vec{a}.vec{b}-15|vec{b}|^2)
c) (8|vec{a}|^2-4vec{a}.vec{b}-15|vec{b}|^2)
d) (|vec{a}|^2+vec{a}.vec{b}-5|vec{b}|^2)
Answer: b
Clarification: To evaluate: ((2vec{a}+5vec{b}).(4vec{a}-5vec{b}))
=(2vec{a}.4vec{a}-2vec{a}.5vec{b}+3vec{b}.4vec{a}-3vec{b}.5vec{b})
=(8|vec{a}|^2+2vec{a}.vec{b}-15|vec{b}|^2)
2. Find the magnitude of (vec{a}) and (vec{b}) which are having the same magnitude and such that the angle between them is 60° and their scalar product is (frac{1}{4}).
a) (|vec{a}|=|vec{b}|=frac{1}{2√2})
b) (|vec{a}|=|vec{b}|=frac{1}{√2})
c) (|vec{a}|=|vec{b}|=frac{1}{2√3})
d) (|vec{a}|=|vec{b}|=frac{2}{√3})
Answer: a
Clarification: Given that: a) (|vec{a}|=|vec{b}|)
b) θ=60°
c) (vec{a}.vec{b}=frac{1}{4})
∴(|vec{a}||vec{b}| cosθ=frac{1}{4})
=(|vec{a}|^2 cos60°=frac{1}{4})
⇒(|vec{a}|^2=frac{1}{4}.frac{1}{2})
∴(|vec{a}|=|vec{b}|=frac{1}{2√2}).
3. If (vec{a}=hat{i}-hat{j}+3hat{k}, ,vec{b}=5hat{i}-2hat{j}+hat{k} ,and ,vec{c}=hat{i}-hat{j}) are such that (vec{a}+μvec{b}) is perpendicular to (vec{c}), then the value of μ.
a) (frac{7}{2})
b) –(frac{7}{2})
c) –(frac{3}{2})
d) (frac{7}{9})
Answer: b
Clarification: Given that: (vec{a}=hat{i}-hat{j}+3hat{k}, ,vec{b}=5hat{i}-2hat{j}+hat{k} ,and ,vec{c}=hat{i}-hat{j})
Also given, (vec{a}+μvec{b}) is perpendicular to (vec{c})
Therefore, ((vec{a}+μvec{b}).vec{c}=0)
i.e. ((hat{i}-hat{j}+3hat{k}+μ(5hat{i}-2hat{j}+hat{k})).(hat{i}-hat{j}))=0
(((1+5μ) ,hat{i}-(1+2μ) ,hat{j}+(μ+3) ,hat{k}).(hat{i}-hat{j}))=0
1+5μ+1+2μ=0
μ=-(frac{7}{2}).
4. Find the angle between (vec{a} ,and ,vec{b}) if (|vec{a}|=2,|vec{b}|=frac{1}{2√3}) and (vec{a}×vec{b}=frac{1}{2}).
a) (frac{2π}{3})
b) (frac{4π}{5})
c) (frac{π}{3})
d) (frac{π}{2})
Answer: c
Clarification: Given that, (|vec{a}|=2, ,|vec{b}|=frac{1}{2√3}) and (vec{a}×vec{b}=frac{1}{2})
We know that, (vec{a}×vec{b}=vec{a}.vec{b} ,sinθ)
∴ (sinθ=frac{(vec{a}×vec{b})}{|vec{a}|.|vec{b}|})
sinθ=(frac{frac{1}{2}}{2×frac{1}{2√3}}=frac{sqrt{3}}{2})
θ=(sin^{-1}frac{sqrt{3}}{2}=frac{π}{3})
5. Find the vector product of the vectors (vec{a}=2hat{i}+4hat{j}) and (vec{b}=3hat{i}-hat{j}+2hat{k}).
a) (hat{i}-19hat{j}-4hat{k})
b) (3hat{i}+19hat{j}-14hat{k})
c) (3hat{i}-19hat{j}-14hat{k})
d) (3hat{i}+5hat{j}+4hat{k})
Answer: c
Clarification: Given that, (vec{a}=2hat{i}+4hat{j}) and (vec{b}=3hat{i}-hat{j}+2hat{k})
Calculating the vector product, we get
(vec{a}×vec{b}=begin{vmatrix}hat{i}&hat{j}&hat{k}\2&4&-5\3&-1&2end{vmatrix})
=(hat{i}(8-5)-hat{j}(4-(-15))+hat{k}(-2-12))
=(3hat{i}-19hat{j}-14hat{k})
6. If (vec{a} ,and ,vec{b}) are two non-zero vectors then ((vec{a}-vec{b})×(vec{a}+vec{b}))=_________
a) (2(vec{a}×vec{b}))
b) ((vec{a}×vec{b}))
c) –(4(vec{a}×vec{b}))
d) (3(vec{a}×vec{b}))
Answer: a
Clarification: Consider ((vec{a}-vec{b})×(vec{a}+vec{b}))
=((vec{a}-vec{b})×vec{a}+(vec{a}+vec{b})×vec{b})
=(vec{a}×vec{a}-vec{b}×vec{a}+vec{a}×vec{b}-vec{b}×vec{b})
We know that, (vec{a}×vec{a}=0,vec{b}×vec{b}=0 ,and ,vec{a}×vec{b}=-vec{b}×vec{a})
∴ (vec{a}×vec{a}-vec{b}×vec{a}+vec{a}×vec{b}-vec{b}×vec{b}=0+2(vec{a}×vec{b})+0)
Hence, ((vec{a}-vec{b})×(vec{a}+vec{b})=2(vec{a}×vec{b}))
7. Find the product ((vec{a}+vec{b}).(7vec{a}-6vec{b})).
a) (2|vec{a}|^2+6vec{a}.vec{b}-3|vec{b}|^2)
b) (8|vec{a}|^2+5vec{a}.vec{b}-5|vec{b}|^2)
c) (2|vec{a}|^2+6vec{a}.vec{b}-8|vec{b}|^2)
d) (7|vec{a}|^2+vec{a}.vec{b}-6|vec{b}|^2)
Answer: d
Clarification: To evaluate: ((vec{a}+vec{b}).(7vec{a}-6vec{b}))
=(vec{a}.7vec{a}-vec{a}.6vec{b}+vec{b}.7vec{a}-6vec{b}.vec{b})
=(7|vec{a}|^2+vec{a}.vec{b}-6|vec{b}|^2)
8. Find the vector product of the vectors (vec{a}=-hat{j}+hat{k}) and (vec{b}=-hat{i}-hat{j}-hat{k}).
a) (2hat{i}-hat{j}+hat{k})
b) (2hat{i}-hat{j}-4hat{k})
c) (hat{i}+hat{j}-hat{k})
d) (2hat{i}-hat{j}-hat{k})
Answer: d
Clarification: Given that, (vec{a}=-hat{j}+hat{k}) and (vec{b}=-hat{i}-hat{j}-hat{k})
Calculating the vector product, we get
(vec{a}×vec{b}=begin{vmatrix}hat{i}&hat{j}&hat{k}\0&-1&1\-1&-1&-1end{vmatrix})
=(hat{i}(1-(-1))-hat{j}(0-(-1))+hat{k}(0-1))
=(2hat{i}-hat{j}-hat{k})
9. If (vec{a}=2hat{i}+3hat{j}+4hat{k}) and (vec{b}=4hat{i}-2hat{j}+3hat{k}). Find (|vec{a}×vec{b}|).
a) (sqrt{685})
b) (sqrt{645})
c) (sqrt{679})
d) (sqrt{689})
Answer: b
Clarification: Given that, (vec{a}=2hat{i}+3hat{j}+4hat{k}) and (vec{b}=4hat{i}-2hat{j}+3hat{k})
∴ (vec{a}×vec{b}=begin{vmatrix}hat{i}&hat{j}&hat{k}\2&3&4\4&-2&3end{vmatrix})
=(hat{i}(9—8)-hat{j}(6-16)+hat{k}(-4-12))
=(17hat{i}+10hat{j}-16hat{k})
∴(|vec{a}×vec{b}|=sqrt{17^2+10^2+(-16)^2})
=(sqrt{289+100+256})
=(sqrt{645})
10. Find the angle between the vectors if (|vec{a}|=|vec{b}|=3sqrt{2}) and (vec{a}.vec{b}=9sqrt{3}).
a) (frac{π}{6})
b) (frac{π}{5})
c) (frac{π}{3})
d) (frac{π}{2})
Answer: a
Clarification: We know that, (vec{a}.vec{b}=|vec{a}|.|vec{b}| ,cosθ)
Given that, (|vec{a}|=|vec{b}|=3sqrt{2} ,and ,vec{a}.vec{b}=9sqrt{3})
(cosθ=frac{vec{a}.vec{b}}{|vec{a}|.|vec{b}|}=frac{9sqrt{3}}{(3sqrt{2})^2}=frac{sqrt{3}}{2})
(θ=cos^{-1}frac{sqrt{3}}{2}=frac{π}{6}).
Mathematics Question Papers for Class 12,