Question :
Find the equation of a line which passes through point A(1,0,-1) and is perpendicular to straight lines $\vec{r} =2 \hat{i}-\hat{j}+\hat{k}+\lambda (2\hat{i}+7\hat{j}-3\hat{k}) $ and $\vec{r} =3\hat{i}-\hat{j}+3\hat{k}+\lambda (2\hat{i}-2\hat{j}+5\hat{k})$
It is given in solution :
Since the line to be determined is perpendicular to the given two straight lines, it is directed towards the vector, ( why so ? )
Can somebody explain one the above statement it will be of great help to me Thanks
Best Answer
Let Direction Cosine of that line be $<a,b,c>\;,$ Then equation of line which passess through
$(1,0,1)$ is $\displaystyle \frac{x-1}{a}=\frac{y-0}{b}=\frac{z-1}{c}.......(1)$
Now above line be Perpendicular to $\vec{r} =2 \hat{i}-\hat{j}+\hat{k}+\lambda (2\hat{i}+7\hat{j}-3\hat{k})$.
So Direction cosine of that line be $<2,7,-3>$
So Dot product of these line be $0.$ So we get $2a+7b-3c=0...............(2)$
Similarly $\vec{r} =3\hat{i}-\hat{j}+3\hat{k}+\lambda (2\hat{i}-2\hat{j}+5\hat{k})$
So Direction of that line be $<2,-2,5>$
So Dot product of $(1)$ line be $2a-2b+5b=0........................................(3)$
Now calculate values of $a,b,c$ from these two equation. and put into $(1).$