The equation of the plane containing the straight line = = and perpendicular to the place containing the straight lines = = and = = is

a. |
x+2y−2z = 0 |
b. |
x−2y+z=0 |

c. |
5x+2y−4z = 0 |
d. |
3x+2y−3z=0 |

The equation of the plane containing the straight line = = and perpendicular to the place containing the straight lines = = and = = is

a. |
x+2y−2z = 0 |
b. |
x−2y+z=0 |

c. |
5x+2y−4z = 0 |
d. |
3x+2y−3z=0 |

1 Answer

127 votes

Vector along the normal to the plane containing the lines

= = and = = is (8 −10 )

Vector perpendicular to the vectors 2 +3 +4 and 8 10 is 26 −52 +26

So, required plane is

26x−52y+26z=0

x−2y+z=0

127 votes

127