First slide

Planes in 3D

Question

$The plane 4 x + 7 y + 4 z + 81 = 0 is rotated through a right angle about its line of intersection with the plane 5 x + 3 y + 10 z = 25 .$
$The equation of the plane in its new position is$

Moderate

A

$x - 4 y + 6 z = 106$
B

$x - 8 y + 13 z = 103$
C

$x - 4 y + 6 z = 110$
D

$x - 8 y + 13 z = 105$

Solution

The equation of the plane through the line of intersection of the planes 4x + 7y + 4z + 81 = 0 and 5x + 3y + 10z = 25 is
$\begin{array}{l} (4 x + 7 y + 4 z + 81) + λ (5 x + 3 y + 10 z - 25) = 0 \\ or (4 + 5 λ) x + (7 + 3 λ) y + (4 + 10 λ) z + 81 - 25 λ = 0 - - - - - (i) \end{array}$
$which is perpendicular to 4 x + 7 y + 4 z + 81 = 0$
$\Rightarrow 4 (4 + 5 λ) + 7 (7 + 3 λ) + 4 (4 + 10 λ) = 0$
$\begin{array}{l} o r 81 λ + 81 = 0 \\ o r λ = - 1 \end{array}$
$Hence, the plane is x - 4 y + 6 z = 106$

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