Question

Let a plane P pass through the point (3, 7, -7) and contain the line, $\frac{x-2}{-3} = \frac{y-3}{2} = \frac{z+2}{1}$. If distance of the plane P from the origin is d, then $d^2$ is equal to _________.

Hint:

To find the equation of a plane containing a point and a line, find two vectors lying in the plane: the direction vector of the line and the vector connecting the given point to a point on the line. The cross product of these two vectors gives the normal to the plane.

Answer 1

Correct Answer: 3

The required plane P contains the point A(3, 7, -7). 
The given line passes through the point B(2, 3, -2) and has a direction vector $\vec{v} = -3\hat{i} + 2\hat{j} + \hat{k}$. 
Since the plane contains both points A and B, the vector $\vec{AB}$ must lie in the plane. 
$\vec{AB} = (2-3)\hat{i} + (3-7)\hat{j} + (-2 - (-7))\hat{k} = -\hat{i} - 4\hat{j} + 5\hat{k}$. 
The normal vector to the plane, $\vec{n}$, must be perpendicular to both $\vec{v}$ and $\vec{AB}$ as both lie in the plane. 

$= \hat{i}(10 - (-4)) - \hat{j}(-15 - (-1)) + \hat{k}(12 - (-2))$ 
$= 14\hat{i} + 14\hat{j} + 14\hat{k} = 14(\hat{i} + \hat{j} + \hat{k})$. 
The direction ratios of the normal are (1, 1, 1). 
The equation of the plane passing through A(3, 7, -7) is $1(x-3) + 1(y-7) + 1(z-(-7)) = 0$. 
$x - 3 + y - 7 + z + 7 = 0 \implies x + y + z - 3 = 0$. 
The distance 'd' of this plane from the origin (0, 0, 0) is given by the formula: 
$d = \frac{|A x_0 + B y_0 + C z_0 + D|}{\sqrt{A^2 + B^2 + C^2}} = \frac{|1(0)+1(0)+1(0)-3|}{\sqrt{1^2+1^2+1^2}} = \frac{|-3|}{\sqrt{3}} = \sqrt{3}$. 
The question asks for the value of $d^2$. 
$d^2 = (\sqrt{3})^2 = 3$.