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If the straight lines $\frac{x-x_1}{l_1}= \frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ and $\frac{x-x_2}{l_2}= \frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$ are coplanar . then how many ways to find the cartesian equation of the plane containing these two lines. I got this question from the 12th public examination in TamilNadu. I answer this in four ways. let $\vec a=x_1\hat i+y_1 \hat j+z_1\hat k $ , $\vec b=x_2\hat i+y_2 \hat j+z_2\hat k $, $\vec c=l_1 \hat i+m_1\hat j+n_1 \hat k$ and $\vec d=l_2 \hat i+m_2\hat j+n_2 \hat k$ the cartesian equation of the plane can be determined by the formula $$[\vec r - \vec a,\vec c,\vec d]=0$$ $$[\vec r-\vec b,\vec c , \vec d]=0$$ $$[\vec r -\vec a , \vec a - \vec b , \vec c]=0$$ $$[\vec r -\vec a , \vec a - \vec b , \vec d]=0$$ are the four possible ways to find the cartesian equation. but, in the answer key, only two ways are given. is my solution correct? if not what is a mistake in it?

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  • $\begingroup$ If $\vec c$ and $\vec d$ are linearly independent, then you can choose any three of the four vectors to form the $4\times4$ determinant-based equation, so up to permutation, I also see $4C3=4$ ways to form it. Aside from the number of ways, how does your solution differ from the answer key? $\endgroup$ – amd Mar 11 '20 at 20:08
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There are two possibilities: either the lines are parallel or they intersect.[Since the question says"THE plane..."we can assume that the lines are distinct.] First we calculate the cross-product $$v=[l_1,m_1,n_1] \times [l_2,m_2,n_2].$$ If $v$ is not the 0 vector,the lines intersect and $v$ is normal to the plane, in which case the equation of the plane can be written as $$v \bullet [x-x_1,y-y_1,z-z_1]=0.$$ If $v$ is the 0 vector, the lines are parallel, in which case we calculate the cross-product $$w=[x_2-x_1,y_2-y_1,z_2-z_1] \times [l_1,m_1,n_1]$$ In this case, $w$ is normal to the plane and the equation of the plane is $$w \bullet [x-x_1,y-y_1,z-z_1]=0.$$.

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