-->

Three Dimensional Geometry

Question
CBSEENMA12033513

Show that the line straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space minus 3 space space straight j with hat on top space plus space 5 space straight k with hat on top space plus space straight lambda space open parentheses straight i with hat on top space minus space straight j with hat on top space plus 2 space straight k with hat on top close parentheses lies in the plane straight r with rightwards arrow on top. space open parentheses 3 straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top close parentheses space plus space 2 space equals 0

Solution
The equation of given plane is
    straight r with rightwards arrow on top. space open parentheses 3 space straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top close parentheses space equals space minus 2
or  straight r with rightwards arrow on top. space open parentheses negative 3 space straight i with hat on top space minus space straight j with hat on top space plus space straight k with hat on top close parentheses space equals space minus 2                                             ...(1)
The equation of given line is
straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 5 space straight k with hat on top space plus space straight lambda open parentheses straight i with hat on top space minus space straight j with hat on top space plus space space 2 space straight k with hat on top close parentheses                          ...(2)
Now line  (2) passes through the point (2, -3, 5), With position vector
2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 5 space straight k with hat on top and is parallel to the vector straight i with hat on top space minus space straight j with hat on top space plus space 2 space straight k with hat on top.
Now plane (1) passes through the point with position vector 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 5 space straight k with hat on top
if   open parentheses 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 5 space straight k with hat on top close parentheses. space space open parentheses negative 3 space straight i with hat on top space minus space straight j with hat on top space plus space straight k with hat on top close parentheses space equals 2
i.e.   if (2) (-3) + (-3) (-1) + (5) (1) = 2
i.e.  if 2 = 2, which is true
Now, negative 3 straight i with hat on top space minus space straight j with hat on top space plus space straight k with hat on top is a vector normal to the plane (1). It will be perpendicular to the line (2) if it is perpendicular straight i with hat on top space minus space straight j with hat on top space plus space 2 space straight k with hat on top       
i.e. if (–3) (1) + (– 1) (– 1) + (1) (2) = 0
i.e. if –3 + 1+ 2 = 0
i.e. if 0 = 0, which is true
∴  line (2) lies in plane (1).

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.