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=60x=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 n1 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)

±4p4p=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  44m+m2=0

 (m2)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^(2.5323)j^(3×2.58)+k^(3343)

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

|P×Q|=14(3i^+j^23k^)