First slide

Planes in 3D

Question

Equation of the plane passing through the points (2,2,1) and (9, 3, 6), and perpendicular to the plane 2x+6y+6z- 1=0 is

Moderate

A

3x + 4y + 5z=9
B

3x + 4y - 5z=9
C

3x - 4y - 5z=9
D

none of the above

Solution

Any plane through (2,2, 1) is a(x-2)+b(y-2)+c(z- 1)=0 -------(i)
It passes through (9, 3, 6) if 7a+ b + 5c=0 --------(ii)
Also (i) is perpendicular to 2x + 6y + 6z - 1 = 0,
we have
$\begin{array}{ll} 2 a + 6 b + 6 c = 0 \\ ∴ a + 3 b + 3 c = 0 - - - - - - (iii) \\ ∴ \frac{a}{- 12} = \frac{b}{- 16} = \frac{c}{20} or \frac{a}{3} = \frac{b}{4} = \frac{c}{- 5} [from (ii) and (iii)] \end{array}$
Therefore, the required plane is 3(x - 2) + 4 (y - 2) - 5 (z- 1) = 0 or 3x + 4y - 5z -9 = 0

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