Question

3i+2j-ak
If vector Rz 21+ 2j- 3k and rector sa
perpendicular vectors then value of 'a' is
given by​

Answer 1

Answer:

$\\\tt -\dfrac{10}{3}$

Explanation:

Since 3i+2j-ak and 2i+2j-3k are perpendicular to each other their dot products must equate to 0.

(3i+2j-ak).(2i+2j-3k )=0

6 + 4 + 3a =0

3a = -10

a =$\\\tt -\dfrac{10}{3}$

Answer 2

Required Answer :-

At first

We know that

$\sf 0 = (3i + 2j + (-ak)) \times (2i+2j+(-3k))$

$\sf 0 = (3i+2j-ak)\times (2i+2j-3k )$

$\sf 0= (3i \times 2i) + (2j \times 2j) + (3k \times ak)$

$\sf 0 = 6+4+ (3k\times ak)$

$\sf 0 = 6+4 + 3a$

$\sf 0=10 + 3a$

$\sf 0 - 10 = 3a$

$\sf -10=3a$

$\sf\dfrac{-10}{3} = a$