Question

The vector $\overrightarrow{c}$ is perpendicular to the vectors $\overrightarrow{a}=(2,-3,1),\overrightarrow{b}=(1,-2,3)$ and satisfies the condition $\overrightarrow{c}$.$(\hat{i}+2\hat{j}-7\hat{k})=10$. Then the vector $\hat{c}$ =

• A. (7,5,1)
• B. (-7,-5,-1)
• C. (1,1,-1)
• D. none

Answer 1

The correct option is A (7,5,1)

Let $\overrightarrow{c}=x\hat{i}+y\hat{j}+z\hat{k}$
Given $\overrightarrow{c}$ is perpendicular to $\overrightarrow{a}$.

Thus, $\overrightarrow{a}.\overrightarrow{c}=0$
$(x\hat{i}+y\hat{j}+z\hat{k}).(2\hat{i}-3\hat{j}+1\hat{k})=0$

$2x-3y+z=0$ --- (1)

Also, $\overrightarrow{c}$ is perpendicular to $\overrightarrow{b}$. Thus, $\overrightarrow{b}.\overrightarrow{c}=0$
$(x\hat{i}+y\hat{j}+z\hat{k}).(\hat{i}-2\hat{j}+3\hat{k})=0$

$x-2y+3z=0$ --- (2)

Finally, $(x\hat{i}+y\hat{j}+z\hat{k}).(\hat{i}+2\hat{j}-7\hat{k})=10$

$x+2y-7z=10$ --- (3)

Solving these three equations, we get $x=7,y=5,z=1$