Question

The plane $3x+4y+6z+7=0$ 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 $x+y+z=0$