The value of $m$ for which straight line $3x-2y+z+3=0=4x-3y+4z+1$ is parallel to the plane $2x-y+mz-2=0$ is
$(A)-2\hspace{1cm}(B)8\hspace{1cm}(C)-18\hspace{1cm}(D)11$
My Attempt:The plane $2x-y+mz-2=0$ is parallel to the $3x-2y+z+3=0$ and $4x-3y+4z+1=0$.
So the lines are perpendicular to the normal of the plane.
So,$(3)(2)+(-2)(-1)+(1)(m)=0$ gives $m=-8$
And $(4)(2)+(-3)(-1)+(4)(m)=0$ gives $m=\frac{-11}{4}$
But the answer given in the book is $m=-2$
There is a method given in the book to answer this question.
Vector $((3\hat{i}-2\hat{j}+\hat{k})\times(4\hat{i}-3\hat{j}+4\hat{k}))$ is perpendicular to $2\hat{i}-\hat{j}+m\hat{k}$
So $m=-2$
But i could not understand this method.Please help me.Thanks.
Best Answer
Note that if you write only $3x-2y+z+3=0$, then it represents a plane.
Here, we have $$3x-2y+z+3=0=4x-3y+4z+1,$$ which represents a line.
We can write the line with the equation $$3x-2y+z+3=0=4x-3y+4z+1$$ as $$\frac{x-0}{1}=\frac{y-\frac{11}{5}}{\frac 85}=\frac{z-\frac 75}{\frac 15}$$ (you can get this by expressing $y,z$ by $x$)
So, $$2\cdot 1+(-1)\cdot \frac 85+m\cdot \frac 15=0\iff m=-2.$$