Question

prove that in Vector product å =3i+6j+9k and b =i-2j+3k are Parallel to each other.​

Answer 1

Explanation:

a = 3i+6j+9k

b = i-2j+3k

a=3[i+2j+3k]=3b

its of the form a=cb

that implies a and b are parallel

Answer 2

Answer:

Not parallel

Explanation:

Given,

$\vec{a} = 3i + 6j+9k$

and $\vec{b} = i - 2j +3k$

As we know that $\vec{a}$ and $\vec{b}$ will be parallel to each other if  and only

       if  $\vec {a}$ ×$\vec{b}$ = 0

Step1:

therefore,

$\vec{a}$ × $\vec{b}$ = $\left[\begin{array}{ccc}i&j&k\\3&6&9\\1&-2&3\end{array}\right]$

         =  $i\left[\begin{array}{cc}6&9\\-2&3\end{array}\right] - j\left[\begin{array}{cc}3&9\\1&3\end{array}\right] +k\left[\begin{array}{cc}3&6\\1&-2\end{array}\right]$

          = i (18+18) -j(9-9)+k(-6 -6)

         =36i -0j -12k

        = 36i - 12 k which is not zero.

Final answer :

Hence, here we see that value of $\vec{a}$ ×$\vec{b}$ not become zero

therefore , given vectors are not parallel to each other.