Question

Find the m if dot product of both vector are perpendicular to each other.

A = 2i + 3 j - 6k and B = 3i - mj + 6k

Answer 1

Explanation:

vector A . vector B =ab cos theta

6-3m-36 = ab cos 90

-3m-30=0

-3m=30

so, m=-10

Answer 2

Answer:

m = - 10

Explanation:

Given :

$\displaystyle \text{ \vec A=2\hat{i}+3\hat{j-6\hat{k}} and }\\\\\displaystyle \text{ \vec B=3\hat{i}-m\hat{j+6\hat{k}} }$

We know formula for dot product

$\displaystyle \vec A.\vec B=AB \ \cos\theta$

when cos θ = cos 90 = 0

Now formula become

$\displaystyle \vec A.\vec B=AB \ 0\\\\\displaystyle \vec A.\vec B=0$

$\displaystyle \text{ \vec A.\vec B=\left(2\hat{i}+3\hat{j-6\hat{k}}\right)\left(3\hat{i}-m\hat{j+6\hat{k}}\right)=0 }\\\\\displaystyle 2\times3-3\times m-6\times6=0\\\\\displaystyle 6-3m-36=0\\\\\displaystyle -3m-30=0\\\\\displaystyle -m-10=0\\\\\displaystyle m=-10$

Thus we get m = - 10.

Extra points :

$\displaystyle \hat{i}\times \hat{i}=1\\\\\displaystyle \hat{j}\times \hat{j}=1\\\\\displaystyle \hat{k}\times \hat{k}=1\\\\\displaystyle \hat{i}\times \hat{k}=0\\\\\displaystyle \hat{j}\times \hat{k}=0\\\\\displaystyle \hat{k}\times \hat{i}=0$