Question

For what value of m the vector $\overrightarrow{A}=2\widehat{i}+3\widehat{j}-6\widehat{k}$ is perpendicular to $\overrightarrow{B}=3\widehat{i}-m\widehat{j}+6\widehat{k}$ .

Answer 1

Hint: As a very first step, read the question well and hence note down the vectors correctly. Now, recall the condition for which the vectors are perpendicular. You could simply take the dot product of the two and then equate it to zero as they are perpendicular to each other and hence find the answer.

Complete step-by-step solution:
In the question, we are given two vectors: $\overrightarrow{A}=2\widehat{i}+3\widehat{j}-6\widehat{k}$and $\overrightarrow{B}=3\widehat{i}-m\widehat{j}+6\widehat{k}$
We are supposed to find the value of m for which the two vectors are perpendicular to each other.
Now, you may recall that when two vectors are perpendicular to each other, their dot product would be zero. That is,
$\overrightarrow{A}\centerdot \overrightarrow{B}=0$
$\Rightarrow \left( 2\widehat{i}+3\widehat{j}-6\widehat{k} \right)\centerdot \left( 3\widehat{i}-m\widehat{j}+6\widehat{k} \right)=0$
On applying dot product here, we get,
$6-3m-36=0$
$\Rightarrow m=\dfrac{36-6}{-3}$
$\therefore m=-10$
Therefore, we found the value of m to be -10.

Note: We have applied the dot product from the known fact that,
$\widehat{i}\centerdot \widehat{i}=\widehat{j}\centerdot \widehat{j}=\widehat{k}\centerdot \widehat{k}=1$
Also, the dot product of two vectors is given by,
$\overrightarrow{A}\centerdot \overrightarrow{B}=\left| A \right|\left| B \right|\cos \theta $
So, here when the angle between the vectors is $90{}^\circ $that is, when the vectors are perpendicular to each other, the cosine of the angle would be zero and so will be the dot product.