Question

The plane $3 x+4 y+6 z+7=0$ is rotated about the line $r =(\hat{ i }+2 \hat{ j }-3 \hat{ k })+t(2 \hat{ i }-3 \hat{ j }+\hat{ k })$ until the plane passes through origin. The equation of the plane in the new position is

• A. x + y + z = 0
• B. 6x + 3y - 4z =0
• C. 4x - 5y - 2z = 0
• D. x + 2y + 4z = 0

Answer 1

Answer: (A)

Equation of plane passes through the origin and containing the line
$r =(\hat{ i }+2 \hat{ j }-3 \hat{ k })+t(2 \hat{ i }-3 \hat{ j }+\hat{ k })$ is
$( r -0) \cdot[(\hat{ i }+2 \hat{ j }-3 \hat{ k }) \times(2 \hat{ i }-3 \hat{ j }+\hat{ k })]=0$
$\Rightarrow r \cdot[\hat{ i }(2-9)-\hat{ j }(1+6)+\hat{ k }(-3-4)]=0$
$\Rightarrow r \cdot(-7 \hat{ i }-7 \hat{ j }-7 \hat{ k })=0$
$\Rightarrow r \cdot(\hat{ i }+\hat{ j }+\hat{ k })=0$
or $x+y+z=0$