Find the coordinates of the point in xz – plane which is the point o

1.	Find the coordinates of the point in xz – plane which is the point of intersection of the plane and the line joining the points (2, 4, 5) and (3, 5,– 4).
Answer» Equation of line joining points (2, 4, 5) and (3, 5, -4) is \(\frac{x-2}{3-2}=\frac{y-4}{5-4}=\frac{z-5}{-4-5}\)----(1) ⇒ \(\frac{x-2}1=\frac{y-4}1 = \frac{z-5}{-9}(Let)\) \(\therefore\) Co-ordinates of arbitrary points of line are (\(\lambda+2,\lambda+4, -9\lambda+5\))----(2) Required point is intersection point of line (1) and xz - plane. \(\therefore\) \(\lambda+4=0\) ⇒ \(\lambda=-4\) (\(\because\) y - co-ordinate of xz-plane is 0) Therefore co-ordinates of required point are (-4 + 2, -4 + 4, -9 x -4 + 5) (From (2)) or (-2, 0, 41). Hence, co-ordinates of required point are (-2, 0, 41)

Find the coordinates of the point in xz – plane which is the point of intersection of the plane and the line joining the points (2, 4, 5) and (3, 5,– 4).

Answer»

Equation of line joining points (2, 4, 5) and (3, 5, -4) is

\(\frac{x-2}{3-2}=\frac{y-4}{5-4}=\frac{z-5}{-4-5}\)----(1)

⇒ \(\frac{x-2}1=\frac{y-4}1 = \frac{z-5}{-9}(Let)\)

\(\therefore\) Co-ordinates of arbitrary points of line are (\(\lambda+2,\lambda+4, -9\lambda+5\))----(2)

Required point is intersection point of line (1) and xz - plane.

\(\therefore\) \(\lambda+4=0\)

⇒ \(\lambda=-4\) (\(\because\) y - co-ordinate of xz-plane is 0)

Therefore co-ordinates of required point are (-4 + 2, -4 + 4, -9 x -4 + 5) (From (2))

or (-2, 0, 41).

Hence, co-ordinates of required point are (-2, 0, 41)

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