Vector

 

Very Short Answer Questions (1 Mark)

Q1. Classify the following measures as scalar and vector quantities:.

(i) 40°

(ii) 50 watt

(iii) 10 gm/cm3

(iv) 20 m/sec towards north

(v) 5 seconds.

A1.

(i) Angle-scalar

(ii) Power-scalar

(iii) Density-scalar

(iv) Velocity-vector

(v) Time-scalar.

Q2. Find the sum of the vectors:

a= i-2j+k, b=-2i+4j+5k and c=i-6j-7k   

A2.

Sum of the vectors = a+b+c

=(i-2j+k)+(-2i+4j+5k)+(i-6j-7k) 

=(i−2i+i) + (-2j+4j−6j)+(k+5k-7k)

=-4j – k.

Q3. What are the horizontal and vertical components of a vector a of magnitude 5 making an angle of 150∘ with the direction of the x-axis?

A3.

The vector a has a magnitude |a|=5 units and makes an angle 

θ=150∘ in the direction of the x-axis. Its horizontal and vertical components can be calculated as follows:

Horizontal component 

x =| a |⋅cosθ

⇒x = 5⋅(cos150∘)

⇒x = 5⋅(−cos30∘)

⇒x = 5⋅(−√3/2)

⇒x = −5√3/2

Vertical component 

y = | a |⋅sinθ

⇒y = 5⋅(sin150∘)

⇒y = 5⋅(sin30∘)

⇒y = 5⋅(1/2)

⇒y = 5/2

Short Answer Questions (4 Mark)

Q1. a , b and c are three mutually perpendicular vectors of equal magnitude. Show that a + b + c makes equal angles with a , b and c with each angle as cos−1(13−−√) .

A1. Let |a→|=|b→|=|c→|=λ and since they are mutually perpendicular,

a→⋅b→=b→⋅c→=a→⋅c→=0 .

Consider |a→+b→+c→|2 :

⇒|a→+b→+c→|2=|a→|2+|b→|2+|c→|2+2(a→⋅b→+b→⋅c→+a→⋅c→)

⇒|a→+b→+c→|2=λ2+λ2+λ2+2(0+0+0)

⇒|a→+b→+c→|2=3λ2

⇒|a→+b→+c→|=3–√λ

Suppose a→,b→,c→ make angles θ1,θ2,θ3 with a→+b→+c→ respectively.

Then, cosθ1=a→⋅(a→+b→+c→)|a→||a→+b+c→|

⇒cosθ1=a→⋅a→+a→⋅b→+a→⋅c→|a→|⋅3–√λ

⇒cosθ1=|a→|2|a→|⋅3–√λ

⇒cosθ1=λ3–√λ

⇒cosθ1=13–√

Similarly, cosθ2=cosθ3=13–√ .

Therefore, θ1=θ2=θ3 .

Q2. If →α =3 ˆi − ˆj and →β =2 ˆi + ˆj +3 ˆk then express →β in the form of →β = →β1 + →β2, where →β1 is parallel to →α and →β2 is perpendicular to →α .

A2. It is given that →α =3 ˆi − ˆj and →β =2 ˆi + ˆj +3 ˆk .

Since →α is parallel to →β1 , that implies →β1 =λ →α .

⇒ →β1 =λ(3 ˆi − ˆj )

Further, →β = →β1 + →β2⇒2 ˆi + ˆj +3 ˆk =[λ(3 ˆi − ˆj )]+ →β2

⇒ →β2 =(2−3λ) ˆi +(1+λ) ˆj +3 ˆk

Now, →β2 is perpendicular to →α , hence their dot product will be equal to zero.

→α ⋅ →β2 =0

⇒(3 ˆi − ˆj )⋅[(2−3λ) ˆi +(1+λ) ˆj +3 ˆk ]=0

⇒3(2−3λ)+(−1−λ)=0

⇒6−9λ−1−λ=0

⇒5−10λ=0

⇒λ=1/2 

We calculate the value of 

→β1 and →β2 :

→β1 =λ(3 ˆi − ˆj )

⇒ →β1=1/2(3 ˆi − ˆj )

⇒ →β1 =3/2ˆi −1/2ˆj

Also, →β2 =(2−3λ) ˆi+(1+λ) ˆj +3 ˆk

⇒ →β2=[2−3(1/2)] ˆi +(1+1/2) ˆj +3 ˆk

⇒ →β2 =1/2ˆi +3/2ˆj +3 ˆk

Hence, we can say β =(3/2ˆi−1/2ˆj )+(1/2ˆi +3/2ˆj +3 ˆk ).

Q3. Let a^,b^,c^ are unit vectors such that a^⋅b^=a^⋅c^=0 and the angle between b^ and c^ is π6 , then prove that a^=±2(b^×c^) .

A3.

 Since a^⋅b^=a^⋅c^=0 , it means that a^

is perpendicular to both b^ and c^ .

This further implies that a^ is perpendicular to the plane in which b^ and c^ lie.

Also, b^×c^=|b^||c^|sinθn^  , where θ is the angle between b^ and c^ , and n^ is a unit vector perpendicular to the plane in which b^ and c^ lie.

⇒b^×c^=(1)(1)sinπ6a^

(Since b^ and c^ are unit vectors, the angle between them is given to be π6 , and a^ satisfies the conditions for n^ )

⇒b^×c^=sinπ6a^

⇒b^×c^=12a^

⇒a^=±2(b^×c^)

Hence proved.

Q4. If a→×b→=c→×d→ and a→×c→=b→×d→ , then prove that a→−d→ is parallel to b→−c→ provided a→≠d→ and b→≠c→ .

A4.

For vectors a→−d→

 and b→−c→

 to be parallel, their cross-product must be equal to zero.

⇒(a→−d→)×(b→−c→)

⇒(a→−d→)×b→−(a→−d→)×c→

⇒a→×b→−d→×b→−a→×c→+d→×c→

⇒a→×b→+b→×d→−a→×c→−c→×d→

 (Since −d→×b→=b→×d→

 and d→×c→=−c→×d→ )

⇒0

 (Since a→×b→=c→×d→ and a→×c→=b→×d→ )

Hence proved.

Q5. Dot product of a vector with vectors i^+j^−3k^ , i^+3j^−2k^ and 2i^+j^+4k^ is 0 , 5 and 8 respectively. Find the vector.

A5. Let the required vector be h→=xi^+yj^+zk^ . We have given a→=i^+j^−3k^ , b→=i^+3j^−2k^ , c→=2i^+j^+4k^ and a→⋅h→=0 ,b→⋅h→=5 , c→⋅h→=8 .

Since a→⋅h→=0 :

(i^+j^−3k^)⋅(xi^+yj^+zk^)=0

⇒x+y−3z=0  ……(1)

Since b→⋅h→=5 :

(i^+3j^−2k^)⋅(xi^+yj^+zk^)=5

⇒x+3y−2z=5 ……(2)

Since c→⋅h→=8 :

(2i^+j^+4k^)⋅(xi^+yj^+zk^)=8

⇒2x+y+4z=8 ……(3)

Subtract equation (1) from (2) :

(x+3y−2z)−(x+y−3z)=5

⇒2y+z=5

⇒y=5−z2

Subtract equation (1) from (3) :

(2x+y+4z)−(x+y−3z)=8

⇒x+7z=8

⇒x=8−7z

Substituting in (1) :

x+y−3z=0

⇒8−7z+5−z2−3z=0

⇒8−10z+5−z2=0

⇒16−20z+5−z2=0

⇒21−21z2=0

⇒21−21z=0

⇒z=1

Now, x=8−7(1)

⇒x=1

Also, y=5−1/2

⇒y=2

Hence, we have h→=i^+2j^+k^.