Q 1 :

For three vectors A→=(-xi^-6j^-2k^), B→=(-i^+4j^+3k^) and C→=(-8i^-j^+3k^), if A→·(B→×C→)=0 then value of x is _______ .             [2024]



(4)       B→=-i^+4j^+3k^ and C→=-8i^-j^+3k^

           B→×C→=|i^j^k^-143-8-13|=15i^-21j^+33k^

           A→·(B→×C→)=(-xi^-6j^-2k^)·(15i^-21j^+33k^)

           0=-15x+126-66

           ⇒15x=60⇒x=4

 



Q 2 :

Two particles are located at equal distance from origin. The position vectors of those are represented by A→=2i^+3nj^+2k^ and B→=2i^–2j^+4pk^, respectively. If both the vectors are at right angle to each other, the value of n–1 is __________.          [2025]



(3)

A→·B→=0 and |A→|=|B→|

⇒ 4 – 6n + 8p = 0

3n – 4p = 2          ... (i)

Also 4+9n2+4=4+4+16p2

3n=±4p          ... (ii)

±4p–4p=2

Taking –ve sign

–8p = 2

p=–14, 3n+1=2 ⇒ n=13



Q 3 :

If two vectors P→=i^+2mj^+mk^ and Q→=4i^–2j^+mk^ are perpendicular to each other. Then, the value of m will be          [2023]

  • –1

     

  • 2

     

  • 3

     

  • 1

     

(2)

P→·Q→=0

(i^+2mj^+mk^)·(4i^–2j^+mk^)=0 ⇒ 4–4m+m2=0

⇒ (m–2)2=0 ⇒ m=2



Q 4 :

Vectors ai^+bj^+k^ and 2i^–3j^+4k^ are perpendicular to each other when 3a + 2b = 7, the ratio of a to b is x2. The value of x is ________ .          [2023]



(1)

For two perpendicular vectors

(ai^+bj^+k^)·(2i^–3j^+4k^)=0

2a – 3b + 4 = 0

on solving, 2a – 3b = –4

also given 3a + 2b = 7

We get a = 1, b = 2

ab=x2 ⇒ x=2ab=2×12=1



Q 5 :

If P→=3i^+3j^+2k^ and Q→=4i^+3j^+2.5k^ then, the unit vector in the direction of P→×Q→ is 1x(3i^+j^–23k^). The value of 'x' is ________ .          [2023]



(4)

P-=3i^+3j^+2k^, Q-=4i^+3j^+2.5k^

P-×Q-=|i^j^k^332432.5|

               =i^(2.53–23)–j^(3×2.5–8)+k^(33–43)

                =0.53i^+0.5j^–3k^=12(3i^+j^–23k^)

|P-×Q-|=14(3i^+j^–23k^)



Q 6 :

When the position vector r→=xi^+yj^+zk^ changes sign as -r→, which one of the following vector will not flip under sign change?                  [2026]

  • Linear momentum

     

  • Velocity

     

  • Acceleration

     

  • Angular momentum

     

(4)

r→=xi^+yj^+zk^

v→=dr→dt=vxi^+vyj^+vzk^

p→=mv→

L→=m(r→×v→)=(xi^+yj^+zk^)×m(vxi^+vyj^+vzk^)

When sign of r→ changes, L→ remains same.