Ans. Given:  

 ∣∣a→∣∣=x2+y2+z2−−−−−−−−−−√=(1)2+(1)2+(1)2−−−−−−−−−−−−−−√=1+1+1−−−−−−−√=3–√|a→|=x2+y2+z2=(1)2+(1)2+(1)2=1+1+1=3

And 

 ∣∣∣b→∣∣∣=x2+y2+z2−−−−−−−−−−√=(2)2+(−7)2+(−3)2−−−−−−−−−−−−−−−−−√=4+49+9−−−−−−−−√=62−−√|b→|=x2+y2+z2=(2)2+(−7)2+(−3)2=4+49+9=62

Also 

 ∣∣c→∣∣=x2+y2+z2−−−−−−−−−−√=(13√)2+(13√)2+(−13√)2−−−−−−−−−−−−−−−−−−−−√=13+13+13−−−−−−−−−√=1|c→|=x2+y2+z2=(13)2+(13)2+(−13)2=13+13+13=1

Ans. Let  and  

Clearly,  [ Coefficients of  and  are same in vectors  and  coefficients of  in  and  are unequal as  ]

But 

And 

Such possible answers are infinite.

Ans. Let  and  =  

  where  = 2 > 0

 Vectors  and  have the same direction.

But 

Such possible vectors are infinite.

Ans. Given:  

If vectors are equal, then their respective components are equal.

Comparing coefficients of  and  on both sides, we have,

 and 

Ans. Let  be the vector with initial point A (2, 1) and terminal point B 

 Position vector of point A is (2, 1) =  and position vector of point B is  =  

  = Position vector of point B – Position vector of point A

= 

= 

  = 

 Scalar components of the vectors  are coefficients of  and  in  i.e.,  and 6 and vector components of the vector  are  and 

Ans. Given:  and  

Adding, 

= 

Ans. We know that a unit vector in the direction of the vector  is 

aˆ=a→∣∣a→∣∣=iˆ+jˆ+2kˆ(1)2+(1)2+(2)2√=iˆ+jˆ+2kˆ1+1+4√=iˆ+jˆ+2kˆ6√a^=a→|a→|=i^+j^+2k^(1)2+(1)2+(2)2=i^+j^+2k^1+1+4=i^+j^+2k^6

 aˆ=16√iˆ+16√jˆ+26√kˆa^=16i^+16j^+26k^

Ans.  Given: Points P (1, 2, 3) and Q (4, 5, 6) 

 Position vector of point P =  =  and position vector of Q =  = , where O is the origin.

  = Position vector of Q – Position vector of P =   

=    = 

Therefore, the unit vector in the direction of vector  = 

= 3iˆ+3jˆ+3kˆ(3)2+(3)2+(3)2√3i^+3j^+3k^(3)2+(3)2+(3)2=

= 

Ans. Given: Vectors  and  

  = 

Therefore, ∣∣∣a→+b→∣∣∣=iˆ+0jˆ+kˆ(1)2+(0)2+(1)2√|a→+b→|=i^+0j^+k^(1)2+(0)2+(1)2

 

Ans. Let  

A vector in the direction of vector  which has magnitude 8 units = 

8.a→∣∣a→∣∣=8(5iˆ−jˆ+2kˆ)(5)2+(−1)2+(2)2√8.a→|a→|=8(5i^−j^+2k^)(5)2+(−1)2+(2)2

⇒⇒    = 

= 

Ans. Let  and  =  =  

  =  where 

Therefore, vectors 2iˆ−3jˆ+4kˆ2i^−3j^+4k^ and −4iˆ+6jˆ−8kˆ−4i^+6j^−8k^ are collinear.

Ans. The given vector is  

  = iˆ+2jˆ+3kˆ(1)2+(2)2+(3)2√=i^+2j^+3k^(1)2+(2)2+(3)2= 

We know that the direction cosines of a vector  are coefficients of  in  i.e., 

Ans. Given: Points A and B 

 Position vector of point A =  = 

And Position vector of point B =  = 

 Vector  =    =    = 

Now  = 

 A unit vector along 

= 

Therefore, the direction cosines of vector  = 

Ans. Let , then ∣∣a→∣∣=(1)2+(1)2+(1)2−−−−−−−−−−−−−−√=3–√|a→|=(1)2+(1)2+(1)2=3 

Let us find angle  (say) between vector  and OX 

  =  = 

 

Similarly angle  (say) between vector  and OY  is 

And angle  (say) between vector  and OZ  is 

  =  = 

 Vector  is equally inclined to OX, OY and OZ.

Ans. Position vector of P is  and Position vector of Q is  

(i) Position vector of point R dividing PQ internally (i.e., R lies within the segment PQ) in the ratio 2 : 1 =  = PR : QR is 

= 

= 

(ii) Position vector of point R dividing PQ externally (i.e., R lies outside the segment PQ and to the right of point Q because ratio 2 : 1 > 1 i.e., PR : QR = 2 : 1) is 

= 

= −3iˆ+3kˆ−3i^+3k^

Ans. Given: Point P (2, 3, 4) and Q 

 Position vector of point P is 

And Position vector of point Q is 

And Position vector of mid-point R of PQ is a→+b→2a→+b→2= 

= 

Ans. Given: Position vector of point A is  

Position vector of point B is  and

Position vector of point C is  where O is the origin.

Now  = Position vector of point B – Position vector of point A

=  = 

=  ……….(i)

 = Position vector of point C – Position vector of point B

=  = 

=  ……….(ii)

 = Position vector of point C – Position vector of point A

=  = 

=  ……….(iii)

Adding eq. (i) and (ii),

 +  =   =  = [By eq. (iii)]

Now From eq. (i),AB = 

From eq. (ii),BC = 

From eq. (iii),AC = 

Here, we can observe that (Longest side BC)² =  AB² + AC²

Also, AB−→−.CA−→−=(+1)(2)+(−3)(−1)+(−5)(1)AB→.CA→=(+1)(2)+(−3)(−1)+(−5)(1) = 2 + 3 – 5 = 0

Thus, AB−→−⊥CA−→−AB→⊥CA→⇒⇒ ∠A=90∘∠A=90∘

Therefore, A B, C are the vertices of a right angled triangle.

Ans. We know by Triangle law of Addition of vectors that    

  

Therefore option (C) is not true because in option (C),

Ans. Option (D) is not true because two collinear vectors can have different directions and also different magnitudes. 

The option (A) and option (C) are true by definition of collinear vectors.

Option (B) is a particular case of option (A) taking