Find the equation of the plane through the poin \( (1,1,1) \) and perp

1.	Find the equation of the plane through the poin \( (1,1,1) \) and perpendicular to the line \( x-2 y+z=2, \quad 4 x+3 x-2 t=0 . \)
Answer» Line is intersection of plane x - 2y + z = 2 and 4x + 3y - z = 0 Let direction ratios of that line are a, b and c. Then that line is perpendicular to both planes. \(\therefore\) a - 2b + c = 0 4a + 3b - c = 0 \(\therefore\) \(\frac{a}{2-3} = \frac {b}{4+1}=\frac{c}{3+8}\) (By cross multiplication method) ⇒ a/-1 = b/5 = c/11 Hence, direction ratios of that line are -1, 5 and 11. Since, line is perpendicular to the required plane. It means line is parallel to the normal of the plane i.e., direction ratios of the normal of the plane are -1, 5 and 11. \(\therefore\) Equation of the plane is -x + 5y + 11z = d----(1) Given the required plane is passing through point (1, 1, 1). \(\therefore\) -1 + 5 + 11 = d ⇒ d = 15 \(\therefore\) Equation of required plane is -x + 5y + 11z = 15.

Find the equation of the plane through the poin \( (1,1,1) \) and perpendicular to the line \( x-2 y+z=2, \quad 4 x+3 x-2 t=0 . \)

Answer»

Line is intersection of plane x - 2y + z = 2 and 4x + 3y - z = 0

Let direction ratios of that line are a, b and c.

Then that line is perpendicular to both planes.

\(\therefore\) a - 2b + c = 0

4a + 3b - c = 0

\(\therefore\) \(\frac{a}{2-3} = \frac {b}{4+1}=\frac{c}{3+8}\) (By cross multiplication method)

⇒ a/-1 = b/5 = c/11

Hence, direction ratios of that line are -1, 5 and 11.

Since, line is perpendicular to the required plane. It means line is parallel to the normal of the plane i.e., direction ratios of the normal of the plane are -1, 5 and 11.

\(\therefore\) Equation of the plane is

-x + 5y + 11z = d----(1)

Given the required plane is passing through point (1, 1, 1).

\(\therefore\) -1 + 5 + 11 = d

⇒ d = 15

\(\therefore\) Equation of required plane is -x + 5y + 11z = 15.

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