Angle between Two Planes

Q. 34. Let ABC be a triangle. Let A be the point (1, 2), y=x is the perpendicular bisector of AB and x-2y+1=0 is the angle bisector of angle C. If the equation of BC is given by ax+by-5=0, then the value of a + b i

Q. The combined equation of angle bisectors between the lines x^2-2xy-3y^2 =0 is

Q. The angle between the planes $r.(2\hat{i}-\hat{j}+2\hat{k})=3$ and $r.(3\hat{i}-6\hat{j}+2\hat{k})=4$

1. $Co{s}^{-1}\left(\frac{16}{21}\right)$
2. $Si{n}^{-1}\left(\frac{4}{21}\right)$
3. $Co{s}^{-1}\left(\frac{1}{4}\right)$
4. $Co{s}^{-1}\left(\frac{3}{4}\right)$

Q.

The plane ax+by =0 is rotated through an angle $\alpha$ about its line of intersection with the plane z=0. Then the equation of the plane in the new position is

1. $ax-by\pm z\sqrt{a^2+b^2}cot\alpha =0$

2. $ax+by\pm z\sqrt{a^2+b^2}tan\alpha =0$

3. $ax-by\pm z\sqrt{a^2+b^2}cot\alpha =0$

4. $ax+by\pm z\sqrt{a^2+b^2}tan\alpha =0$

Q. 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

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

Q. The angle between the planes $2x+y-2z+3=0$ and $6x+3y+2z=5$ is

1. ${\cos }^{-1}\left(\frac{7}{11}\right)$
2. ${\cos }^{-1}\left(\frac{9}{21}\right)$
3. ${\cos }^{-1}\left(\frac{11}{21}\right)$
4. ${\cos }^{-1}\left(\frac{13}{29}\right)$