Question

In order to find the dip of oil bed below the surface of the ground, vertical boring are made from angular points $A,B,C$ of a $△ABC$ which is in a horizontal plane. Let the depth of oil bed at three points $A,B,C$ are found to be $l,$ $l+k$ and $l+m$ $(k<m)$ respectively. The length of the sides $CA$ and $AB$ are $b$ and $c$ respectively and the angle between them is $A$. If the angle of the dip with horizontal is $\theta$, then

• A. $\tan \theta \sin A=\sqrt{\frac{k^2}{c^2}+\frac{m^2}{b^2}-\frac{2km}{bc}\cos A}$
• B. $\tan \theta \sin A=\sqrt{\frac{k^2}{b^2}+\frac{m^2}{c^2}-\frac{2km}{bc}\cos A}$
• C. Normal to the oil bed plane is $\overrightarrow{n}=bm\hat{i}+ck\hat{j}+bc\hat{k}$, when $A=\frac{\pi }{2}$
• D. Normal to the oil bed plane is $\overrightarrow{n}=mc\hat{i}+bk\hat{j}+bc\hat{k}$, when $A=\frac{\pi }{2}$

Answer 1

Answer: (A), (D)

Normal to the oil bed plane is $\overrightarrow{n}=mc\hat{i}+bk\hat{j}+bc\hat{k}$, when $A=\frac{\pi }{2}$


Oil bed is shown by $A^{\prime }PQ$.
$\theta$ is the angle between plane $A^{\prime }PQ$ and $A^{\prime }{B}^{\prime }{C}^{\prime }$.
Let $A^{\prime }{B}^{\prime }{C}^{\prime }$ be $xy$ -plane with $x$- axis along $A^{\prime }{C}^{\prime }$ and $A^{\prime }$ be the origin.
Position vectors of
$C^{\prime }:b\hat{i}{B}^{\prime }:c\cos A\hat{i}+c\sin A\hat{j}Q:b\hat{i}-m\hat{k}P:c\cos A\hat{i}+c\sin A\hat{j}-k\hat{k}$
Unit normal vector to $A^{\prime }{B}^{\prime }{C}^{\prime }$ is,
${\hat{n}}_1=\hat{k}$
Normal vector to plane $A^{\prime }PQ$,
${\overrightarrow{n}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ b& 0& -m\\ c\cos A& c\sin A& -k\end{array}\right|=\left(mc\sin A\right)\hat{i}+(bk-mc\cos A)\hat{j}+\left(bc\sin A\right)\hat{k}$
When $A=\frac{\pi }{2}$,
Normal is,
${\overrightarrow{n}}_2=mc\hat{i}+bk\hat{j}+bc\hat{k}$

The angle between plane $A^{\prime }PQ$ and $A^{\prime }{B}^{\prime }{C}^{\prime }$,
$\cos \theta =\frac{|{\hat{n}}_1\cdot {\overrightarrow{n}}_2|}{|{\overrightarrow{n}}_2|}\Rightarrow \cos \theta =\frac{|bc\sin A|}{\sqrt{(mc\sin A{)}^2+(bk-mc\cos A{)}^2+(bc\sin A{)}^2}}\Rightarrow \cos \theta =\frac{bc\sin A}{\sqrt{b^2{c}^2{\sin }^{2}A+(b^2{k}^2+m^2{c}^2-2bcmk\cos A)}}\therefore \tan \theta =\frac{\sqrt{b^2{k}^2+m^2{c}^2-2bcmk\cos A}}{bc\sin A}$
Hence,
$\tan \theta \sin A=\sqrt{\frac{k^2}{c^2}+\frac{m^2}{b^2}-\frac{2km}{bc}\cos A}$