Question

vector 2i+3j-7k and i+mj-2k are perpendicular vectoers the value of m is​

Answer 1

Question:

Vector $2 \hat{i}+3 \hat{j} -7 \hat{k}$ and $\hat{i}+m \hat{j}-2 \hat{k}$ are perpendicular vectors, the value of m is:

Solution:

We will use the idea that, when two vectors are perpendicular, their dot product is always 0.

Let's assume two vectors as $\vec{a}$ and $\vec{b}$ .

$\vec{a}.\vec{b}= ab~cos \theta ~ [Taking \theta = 90°] \\\\ \vec{a}.\vec{b}= ab~cos90° \implies 0$

So,as per the question,

$( 2 \hat{i}+3 \hat{j} -7 \hat{k}). (\hat{i}+m \hat{j}-2 \hat{k}) =0 \\\\ 2 \hat{i}.\hat{i} + (3m)\hat{j}.\hat{j} + (-7 \times - 2) \hat{k}.\hat{k} =0\\\\ 2 + 3m+14=0 \\\\ 3m = - 16 \\\\ \boxed{\bf m =\frac{-16}{3}}$

More to know:

Dot product of an unit vector with itself is 1

$\hat{i}.\hat{i}=1 \\\\ \hat{j}.\hat{j}=1 \\\\ \hat{k}.\hat{k}=1$